Moving Charges & Magnetism

1Given

• One long wire carrying a steady current I.
• First shape: a circular loop of one turn. Field at its centre = B.
• Second shape: the same wire bent into a coil of n turns.

The word “bent” is the hidden data: nobody cuts the wire or adds to it, so its total length is the same in both shapes.

2Asked

The magnetic field at the centre of the n-turn coil, in terms of B.

3Concept & formula

Each turn of a coil contributes its own field at the centre, and all n turns push the same way, so n multiplies the single-turn value.

4Steps

1. Call the first radius R. Its circumference is the whole wire, so the wire length is L = 2πR.
2. Field of that single turn: B = μ₀(1)I / 2R = μ₀I / 2R.  (equation 1)
3. Now bend it into n turns, each of radius r. The n circles together use up the same wire: n × 2πr = L.
4. Substitute L = 2πR:   n × 2πr = 2πR.
5. Cancel 2π from both sides: n r = R.
6. So r = R / n. Each new circle is n times smaller. This is the step most students miss.
7. Field of the new coil: B′ = μ₀ n I / 2r.
8. Put r = R/n into it: B′ = μ₀ n I / (2 × R/n).
9. Dividing by R/n is the same as multiplying by n/R: B′ = μ₀ n I × n / 2R = μ₀ n² I / 2R.
10. Compare with equation 1: the whole thing is n² × (μ₀I / 2R) = n²B.

5Shortcut

Two gains multiply: n turns (×n) and each turn n times closer to the centre (×n). Total ×n². Any “same wire re-bent” question gives a square.

If instead a fresh longer wire made n turns of the same radius R, the answer would be only nB.

1Given

• An electron moving in a circular path.
• Transverse magnetic field, B = 3.57 × 10⁻² T. (“Transverse” = at 90° to the velocity.)
• Specific charge, e/m = 1.76 × 10¹¹ C/kg.

Not given: the speed and the radius. So the answer must not need them.

2Asked

The frequency of revolution — how many complete circles the electron makes per second.

3Concept & formula

The magnetic force always acts sideways, so it cannot change the speed — only the direction. It supplies the centripetal force.

4Steps

1. Force needed to keep a body of mass m moving in a circle of radius r at speed v: F = mv²/r.
2. Force supplied by the field: F = evB sin 90° = evB.
3. Equate them: evB = mv²/r.
4. Both sides carry a v, so cancel one: eB = mv / r.
5. Rearrange for the radius: r = mv / eB.
6. One lap is a distance 2πr covered at speed v, so the time for one lap is T = 2πr / v.
7. Put r = mv/eB into that: T = (2π / v) × (mv / eB).
8. The v cancels: T = 2πm / eB. The speed has vanished — that is why it was never given.
9. Frequency is one over the period: f = eB / 2πm.
10. Group the constants the way the question supplies them: f = (e/m) × B / 2π.
11. Substitute: f = (1.76 × 10¹¹)(3.57 × 10⁻²) / (2π).
12. Multiply the front numbers: 1.76 × 3.57 = 6.2832. Powers: 10¹¹ × 10⁻² = 10⁹. Top = 6.2832 × 10⁹.
13. Bottom: 2π = 6.2832.
14. Divide: f = 6.2832 × 10⁹ / 6.2832 = 1 × 10⁹ Hz = 1 GHz.

5Shortcut

The setter chose 1.76 × 3.57 = 6.2832 = 2π exactly. Whenever a clean 2π appears on top, the answer is a bare power of ten. Spot it and skip the division.

1Given

• Long straight wire XY carrying current I.
• Square loop ABCD of side L, carrying current i, clockwise as drawn.
• Coplanar with the wire (flat in the same plane).
• Near side AB is at distance L/2 from the wire.

2Asked

The net force on the loop.

3Concept & formula

The wire's field weakens with distance, so the near side sits in a stronger field than the far side. Both feel a force; the two forces oppose, so subtract.

4Steps

1. Distance of the near side AB from the wire: L/2 (given).
2. Distance of the far side CD: L/2 + L = 3L/2, because the square's side is L.
3. Deal with the top and bottom sides BC and AD first. They are mirror images about the mid-line, carrying current in opposite directions through the same field, so their forces are equal and opposite. They cancel exactly. Ignore them.
4. Field at AB: B₁ = μ₀I / (2π × L/2). Dividing by L/2 means multiplying by 2, so B₁ = μ₀I / πL.
5. Angle check for AB: its current runs parallel to the wire, while the field there points into or out of the page. Perpendicular, so sin θ = 1.
6. Force on AB: F₁ = B₁ × i × L = (μ₀I / πL) × i × L. The L cancels: F₁ = μ₀Ii / π.
7. Field at CD: B₂ = μ₀I / (2π × 3L/2) = μ₀I / 3πL.
8. Force on CD: F₂ = B₂ × i × L = μ₀Ii / 3π.
9. Directions: in AB the current runs the same way as I, so they attract — F₁ pulls the loop toward the wire. In CD the current runs opposite to I, so they repel — F₂ pushes it away.
10. Opposite directions means subtract: F = F₁ − F₂ = μ₀Ii/π − μ₀Ii/3π.
11. Common denominator 3π: F = (3μ₀Ii − μ₀Ii) / 3π = 2μ₀Ii / 3π, pointing toward the wire.

5Shortcut

L cancels out, so any option containing L is wrong on sight — that removes two choices immediately. And a loop near a straight wire is always pulled in, because the near side always sits in the stronger field.

1Given

• A straight wire of radius a — a solid rod, not a thin wire.
• Steady current I, uniformly distributed over the cross-section.
• Point 1 at r = a/2, which is inside the metal.
• Point 2 at r = 2a, which is outside the wire.

2Asked

The ratio B(at a/2) : B′(at 2a).

3Concept & formula

Two different regions need two different formulas. Ampère's law asks only one question: how much current does my circle enclose?

4Steps

1. Draw an imaginary circle of radius r about the wire's axis. By symmetry B has the same size all around it, so ∮B·dl = B × 2πr.
2. Ampère's law: B × 2πr = μ₀ × (current inside that circle).
3. Inside the metal the circle catches only part of the current. Current is spread evenly, so the share of current equals the share of area.
4. Share of area = πr² / πa² = r² / a². Hence I(enc) = I r² / a².
5. Check at r = a/2: I(enc) = I (a/2)² / a² = I(a²/4)/a² = I/4. Half the radius gives a quarter the current, because area goes as r².
6. Insert into Ampère: B × 2πr = μ₀ I r² / a², so B = μ₀ I r / 2πa².
7. Put r = a/2: B = μ₀I(a/2) / 2πa² = μ₀I / 4πa.
8. Outside the wire the circle now catches all the current: I(enc) = I.
9. So B′ × 2πr = μ₀I, giving B′ = μ₀I / 2πr.
10. Put r = 2a: B′ = μ₀I / (2π × 2a) = μ₀I / 4πa.
11. The two results are identical, so the ratio is 1 : 1.

5Shortcut

Inside, B ∝ r/a². Outside, B ∝ 1/r. Compare (a/2)/a² with 1/(2a) — both equal 1/2a. One line, no constants needed.

Remember the shape: B rises straight from zero at the centre, peaks at the surface r = a, then falls as 1/r.

1Given

• A wire carrying current I, made of three parts.
• Two long straight portions, parallel to the x-axis, each running off to infinity.
• A semicircular portion of radius R lying in the yz-plane, centred at O.
• The straight parts join the two ends of the semicircle, so each sits at perpendicular distance R from O.

2Asked

The magnetic field at point O, in unit-vector form.

3Concept & formula

Fields add as vectors. Work out each of the three pieces separately, then add.

4Steps

1. A full circular loop gives μ₀I/2R at its centre.
2. A semicircle is exactly half of it: B = ½ × μ₀I/2R = μ₀I / 4R.
3. Direction of the semicircle's field: a loop's field at its centre is perpendicular to the loop's plane. The plane is yz, so the field is along the x-axis. Curling the right hand along the current gives the thumb along −x. So this part is −(μ₀I/4R) î.
4. An infinite straight wire gives μ₀I/2πd. Each straight part here is only a half wire, starting at the semicircle's end and running to infinity, with O on the perpendicular through that end.
5. So each gives half: B = ½ × μ₀I/2πR = μ₀I / 4πR.
6. Direction of each straight part's field: the wires run along x and lie in the ±y direction from O, so the field must be perpendicular to both — along z. Applying the right-hand rule to each wire separately, both come out along −k̂.
7. Same direction means they add: total from the two straights = 2 × μ₀I/4πR, along −k̂.
8. Sum the two contributions: B = −(μ₀I/4R) î − (2μ₀I/4πR) k̂.
9. To combine, give both terms the denominator 4πR. Multiply the first term top and bottom by π: μ₀I/4R = πμ₀I / 4πR.
10. Factor out the common −μ₀I/4πR: B = −(μ₀I / 4πR)(π î + 2 k̂).

5Shortcut

All four options share the same magnitude, so only two decisions matter. The π can only come from the circular part (it appears when μ₀I/4R is written over 4πR), and the 2 can only come from two identical straight halves adding. Both fields are negative, so the two terms carry the same sign inside the bracket — that rules out the mixed-sign options at once.

1Given

• One electron, charge of magnitude e.
• Circular orbit of radius r.
• It completes n rotations every second.

There is no wire and no battery here — just one moving charge.

2Asked

The magnitude of the magnetic field produced at the centre.

3Concept & formula

Current means charge passing a fixed point per second. A charge going round and round is therefore a current, and a circular path of current is just a one-turn loop.

4Steps

1. Stand at one point on the orbit and watch. Each lap, the electron passes you once, carrying charge e.
2. It does this n times per second, so the charge passing per second is n × e.
3. Therefore I = ne.
4. Longer route, same answer: the time for one rotation is T = 1/n, so I = q/T = e/(1/n) = ne.
5. The path is a circle, so treat it as a single circular loop of radius r carrying current I.
6. Field at the centre of a circular loop: B = μ₀I / 2r.
7. Substitute I = ne: B = μ₀(ne) / 2r = μ₀ne / 2r.

5Shortcut

Circle → divide by 2r. Straight wire → divide by 2πd. An option carrying a stray π is the straight-wire trap.

If the speed v is given instead of n, use I = ev/2πr, which leads to B = μ₀ev / 4πr².

1Given

• A loop carrying current i, centred on O, made of four pieces.
• Arc I: radius R, covering 240° (two-thirds of a circle).
• Arc II: radius 2R, covering 120° (one-third of a circle).
• Two straight pieces joining the arcs, both lying along radii through O.

2Asked

The magnetic field at the central point O, size and direction.

3Concept & formula

An arc gives the same fraction of a full loop's field as the fraction of the circle it covers. A straight piece pointing at O gives nothing.

4Steps

1. Take the straight pieces first. Both lie along radii, so the current direction and the line to O are along the same line.
2. The Biot–Savart contribution goes as dl × r̂, and the angle between them is 0°. Since sin 0° = 0, both straight pieces contribute exactly zero. Delete them.
3. Arc I covers 240° out of 360°, which is 240/360 = 2/3 of a circle.
4. A full circle of radius R would give μ₀i/2R, so arc I gives B₁ = (2/3) × μ₀i/2R = 2μ₀i/6R = μ₀i / 3R.
5. Arc II covers 120°, which is 120/360 = 1/3 of a circle.
6. Careful — arc II has radius 2R, so a full circle of that radius would give μ₀i/(2 × 2R) = μ₀i/4R.
7. Arc II therefore gives B₂ = (1/3) × μ₀i/4R = μ₀i / 12R.
8. Walk the current around the loop: both arcs circle O in the same sense, so both fields point the same way. Add them.
9. Common denominator 12: B = μ₀i/3R + μ₀i/12R = 4μ₀i/12R + μ₀i/12R.
10. Total B = 5μ₀i / 12R.
11. Direction: point the right thumb along the current and curl. For the sense drawn, the field at O points into the page, described in the options as downward.

5Shortcut

Delete radial straight pieces on sight — that is usually half the work gone.

If the drawn angles are unclear, use the closure rule θ₁ + θ₂ = 360° and test the likely splits against the printed options. A 270°/90° reading here would give 7μ₀i/16R, which is absent, confirming 240°/120° was intended.

1Given

• Proton speed v = 3 × 10⁶ m/s, moving horizontally from east to west.
• Uniform magnetic field B = 2 × 10⁻³ T, pointing vertically upward.
• Not printed, but expected knowledge: q = 1.6 × 10⁻¹⁹ C and m = 1.67 × 10⁻²⁷ kg.

2Asked

The acceleration of the proton.

3Concept & formula

Find the magnetic force, then divide by the mass. Nothing else is happening.

4Steps

1. Check the angle first. The velocity is horizontal (east to west) and the field is vertical (upward). Horizontal and vertical are perpendicular, so θ = 90° and sin θ = 1.
2. Force: F = qvB = (1.6 × 10⁻¹⁹)(3 × 10⁶)(2 × 10⁻³).
3. Multiply the front numbers: 1.6 × 3 × 2 = 9.6.
4. Add the powers: (−19) + 6 + (−3) = −16.
5. So F = 9.6 × 10⁻¹⁶ N.
6. Acceleration: a = F/m = 9.6 × 10⁻¹⁶ / 1.67 × 10⁻²⁷.
7. Divide the front numbers: 9.6 ÷ 1.67 ≈ 5.75.
8. Subtract the powers: −16 − (−27) = −16 + 27 = +11.
9. So a ≈ 5.8 × 10¹¹ m/s².
10. Direction, for completeness: v is west, B is up, and the charge is positive, so F = qv × B points south. The proton curves sideways; its speed never changes.

5Shortcut

The wrong options are exactly 2×, 3× and ½× the right answer, so an arithmetic slip cannot rescue you — get the constants right. Also note option (b) is printed in m/s, not m/s²; a wrong-dimension option can be struck off on sight.

1Given

• A circular coil carrying current I ampere; field at its centre is B.
• The coil is bent into a smaller circular coil of n turns; new field is B′.

“Bent” and “smaller” are both telling you the same thing: the wire is unchanged in length, so the radius must shrink.

2Asked

The ratio B′ : B.

3Concept & formula

4Steps

1. Original coil: one turn of radius R, so the wire length is L = 2πR and B = μ₀I / 2R.
2. New coil: n turns of radius r, using the same wire, so n(2πr) = 2πR.
3. Cancel 2π: nr = R, hence r = R/n.
4. New field: B′ = μ₀ n I / 2r = μ₀ n I / (2R/n).
5. Multiply top and bottom out: B′ = μ₀ n² I / 2R.
6. Divide by the original: B′/B = (μ₀n²I/2R) ÷ (μ₀I/2R).
7. Everything cancels except n²: B′/B = n².

5Shortcut

Identical to Q1. Read for the word “bent”: if it's there, the answer carries a square. Option n : 1 is placed for those who miss it.

1Given

• Two very long, straight, parallel wires.
• Equal currents, both flowing the same way.

2Asked

Whether they attract, repel, or neither.

3Concept & formula

Each wire creates a field at the other wire's location, and each then feels a force in that field.

4Steps

1. Suppose both currents flow upward, with wire 2 to the right of wire 1.
2. Field of wire 1 at the position of wire 2: point the right thumb up and curl. On the right-hand side the fingers point into the page.
3. Force on wire 2: F = I L × B. Its current is up, the field is into the page. Up crossed with into-the-page points to the left — back toward wire 1.
4. Repeat for wire 1 sitting in wire 2's field: field is out of the page there, and the force comes out pointing right, toward wire 2.
5. Each wire is pulled toward the other, so they attract.
6. Magnitude, per unit length: F/L = μ₀I₁I₂ / 2πd. This relation is the SI definition of the ampere.

5Shortcut

Like currents attract, unlike currents repel — the exact opposite of charges. Keep the two rules in separate drawers.

Intuition: two parallel currents flowing the same way are just one thicker current seen up close, and a current always wraps its field around itself, squeezing inward.

1Given

• A bell-shaped graph of B against distance X along the axis of a circular coil.
• Two marked points A and A′, sitting symmetrically on the two flanks.

2Asked

Which of the four statements about A and A′ is false.

3Concept & formula

A and A′ are the points of inflection, where the curve changes from bending one way to bending the other.

4Steps

1. Statement (a), “points of zero curvature”: curvature is measured by the second derivative, which passes through zero at an inflection point. True.
2. Statement (b), “B varies linearly with X there”: zoom in on A and the curve is indistinguishable from a straight line. That local straightness is the whole point of an inflection. True.
3. Statement (c), “dB/dx = 0 at A and A′”: look at the dashed tangents in the figure. They are steep. In fact the slope reaches its maximum magnitude at the inflection points. False.
4. The slope is zero only at the peak, x = 0, where the field is largest.
5. Statement (d), “d²B/dx² = 0 at A and A′”: that is the definition of an inflection point. True.
6. So (c) is the false statement.

5Shortcut

First derivative zero means flat. Second derivative zero means straight. A steep ramp is perfectly straight but nowhere near flat — that single distinction answers the question.

This is also the principle behind Helmholtz coils: place two coils so that their inflection points coincide, and the two locally straight parts cancel, giving a uniform field.

The paper prints dB/dt; the graph's axis is X, so read it as dB/dx.

1Given

• Two particles A and B with equal charges.
• Both accelerated through the same potential difference V.
• Both then enter the same uniform magnetic field B.
• They describe circles of radii R₁ and R₂.

2Asked

The ratio of the mass of A to the mass of B.

3Concept & formula

Two stages: an electric stage that sets the speed, then a magnetic stage that sets the radius.

4Steps

1. Acceleration stage: the work done by the electric field becomes kinetic energy, qV = ½mv².
2. Solve for the speed: v² = 2qV/m, so v = √(2qV/m).
3. Read that carefully — m is under the root in the denominator, so a heavier particle emerges slower.
4. Magnetic stage: r = mv / qB.
5. Substitute the speed: r = (m/qB) × √(2qV/m).
6. To simplify, write the leading m as √(m²) so everything sits under one root: r = (1/qB) × √(m² × 2qV/m).
7. Inside the root, m²/m = m, leaving √(2mqV).
8. So r = √(2mqV) / qB, which tidies to r = (1/B)√(2mV/q).
9. For these two particles q, V and B are identical, so everything except m is common. Hence r ∝ √m.
10. Therefore R₁/R₂ = √(m₁/m₂).
11. Square both sides: m₁/m₂ = (R₁/R₂)².

5Shortcut

Stopping at r = mv/qB and saying “r ∝ m” gives the wrong option. Two effects fight: more mass means more momentum (raises r) but also a lower speed from the same voltage (lowers r). The square root is what survives.

Rule: accelerated through a potential difference always means a square root is coming.

1Given

• A ring of radius r.
• Linear charge density λ, meaning charge per unit length in C/m.
• Rotating uniformly with angular velocity ω.

2Asked

The magnetic field produced at the centre of the ring.

3Concept & formula

A spinning ring of charge is a current loop. Convert λ and ω into a current first, then apply the loop formula.

4Steps

1. Total charge on the ring = density × length. The ring's length is its circumference, 2πr.
2. So Q = λ(2πr).
3. Rotations per second: from ω = 2πf, we get f = ω/2π.
4. Stand at one point on the ring. Every rotation the whole charge Q sweeps past you, and this happens f times a second.
5. So I = Qf = λ(2πr) × (ω/2π).
6. The 2π cancels: I = λ r ω.
7. Field at the centre of a loop: B = μ₀I / 2r = μ₀(λrω) / 2r.
8. The r cancels: B = μ₀λω / 2.

5Shortcut

r vanishing makes physical sense: a bigger ring carries more charge moving faster (current ∝ r) but sits further from the centre (field ∝ 1/r). The two effects cancel exactly.

Three options can be killed by inspection: one has no μ₀, one squares λ, and one puts ω in the denominator (which would mean a faster spin gives a weaker field).

1Given

• Two particles A and B, equal charges of +6 C each.
• Accelerated through the same potential difference.
• Same magnetic field; radii 2 cm and 3 cm respectively.

2Asked

The ratio of the mass of A to that of B.

3Concept & formula

4Steps

1. Same charge, same voltage, same field, so every factor except m is shared between the two particles.
2. From r = (1/B)√(2mV/q), the only surviving dependence is r ∝ √m.
3. Write the ratio: R(A)/R(B) = √(m(A)/m(B)).
4. Square both sides: m(A)/m(B) = (R(A)/R(B))².
5. Substitute the numbers: (2/3)².
6. Both radii are in the same unit, so no conversion to metres is needed — the units cancel inside the ratio.
7. (2/3)² = 4/9.

5Shortcut

The +6 C is a decoy; equal charges always cancel. The trap option is the un-squared 2/3, for anyone who writes r ∝ m.

Turn it around: if the masses were equal and the charges differed, then r ∝ 1/√q, and the charge ratio would be the inverse square of the radius ratio.

1Given

• Magnetic field 10⁻⁴ Wb/m². Wb/m² is another name for the tesla, so B = 10⁻⁴ T.
• The beam enters normally, meaning perpendicular, so sin θ = 1.
• Specific charge q/m = 10¹¹ C/kg, given as one combined quantity.
• Velocity v = 10⁹ m/s.

2Asked

The radius of the circular path.

3Concept & formula

Rewrite the radius formula so that the block q/m stays intact — don't try to split it.

4Steps

1. Start from r = mv / qB.
2. Divide top and bottom by m: r = v / [(q/m) × B].
3. Substitute: r = 10⁹ / [(10¹¹)(10⁻⁴)].
4. Work out the denominator: 10¹¹ × 10⁻⁴ = 10^(11−4) = 10⁷.
5. Divide: r = 10⁹ / 10⁷ = 10^(9−7) = 10².
6. So r = 100 m.

5Shortcut

Whenever q/m is handed to you as one number, use r = v/[(q/m)B] and f = (q/m)B/2π. These two cover almost every “specific charge” question.

A physics aside, not needed for the mark: 10⁹ m/s is faster than light, so this particle cannot exist. Exam setters pick round numbers without checking.

1Given

• A cyclotron: two hollow D-shaped boxes in a strong magnetic field, with an alternating voltage across the gap between them.

2Asked

Which kinds of particle it can accelerate.

3Concept & formula

Inside a dee the magnetic field bends the particle into a semicircle. In the gap the electric field speeds it up. Both stages need charge.

4Steps

1. Write both forces and look at what they share: every term contains q.
2. If q = 0, both forces are zero. A neutron would sail straight through, never curving and never accelerating. So neutrons are out.
3. Does the sign of q matter? A negative charge feels the force the other way, so it simply circles in the opposite sense. Flip the field or the phase of the voltage and the machine works exactly the same.
4. So “only positive” and “only negative” are both wrong.
5. Therefore both positively and negatively charged particles can be accelerated.
6. Worth knowing: electrons are unsuitable in practice, but because of their tiny mass, not their sign. From f = qB/2πm, a small m makes f impractically high, and electrons turn relativistic almost at once, so m rises, f drifts, and they fall out of step with the voltage.

5Shortcut

No charge means no electric or magnetic force at all — neutrons, neutrinos and photons cannot be accelerated by any such machine. That single fact answers most cyclotron conceptual questions.

Keep ready: f = qB/2πm and K(max) = q²B²R²/2m.

1Given

• Two parallel beams of positrons (positive charges).
• Both moving in the same direction.

2Asked

Whether the beams attract, repel, or neither.

3Concept & formula

Two forces act at once, and they oppose each other. The answer depends on which is bigger.

4Steps

1. A beam of positrons is a stream of moving charges, so it is both charged and a current.
2. Electric effect: the beams are both positive, so they repel.
3. Magnetic effect: parallel currents in the same direction attract.
4. Compare their sizes. For two lines of charge with density λ moving at speed v, the current is I = λv.
5. Electric force per unit length: F(E) = λ² / 2πε₀d.
6. Magnetic force per unit length: F(B) = μ₀(λv)² / 2πd.
7. Divide: F(B)/F(E) = μ₀ε₀v².
8. Use c = 1/√(μ₀ε₀), which means μ₀ε₀ = 1/c².
9. So F(B)/F(E) = v²/c².
10. Since v is always less than c, this ratio is always less than 1. The magnetic attraction is always weaker than the electric repulsion.
11. Net effect: the beams repel.

5Shortcut

Why doesn't this contradict Q10? Because a wire is electrically neutral — the moving electrons are balanced by the fixed positive lattice, so only the magnetic force survives. A free beam has no neutralising background, so its naked charge wins.

Rule: wires attract, beams repel.

1Given

• Two coils, each of radius R = 2π cm = 2π × 10⁻² m.
• Concentric — they share the same centre.
• Their planes are at right angles to each other.
• Currents 3 A and 4 A.

2Asked

The magnetic induction at the common centre, in Wb/m².

3Concept & formula

A loop's field at its centre points perpendicular to its own plane. Planes at 90° therefore give fields at 90°, and perpendicular vectors add by Pythagoras.

4Steps

1. Field of coil 1: B₁ = μ₀(3) / 2R.
2. Field of coil 2: B₂ = μ₀(4) / 2R.
3. Both share the factor μ₀/2R, so factor it out: B(net) = (μ₀/2R)√(3² + 4²).
4. 3² + 4² = 9 + 16 = 25, and √25 = 5. The 3-4-5 triangle is the giveaway that Pythagoras was intended.
5. So B = 5μ₀ / 2R.
6. Substitute μ₀ = 4π × 10⁻⁷ and R = 2π × 10⁻²: B = (4π × 10⁻⁷)(5) / [2 × 2π × 10⁻²].
7. The denominator is 4π × 10⁻².
8. The 4π cancels top and bottom: B = 5 × 10⁻⁷ / 10⁻².
9. Subtract the powers: −7 − (−2) = −5. So B = 5 × 10⁻⁵ Wb/m².

5Shortcut

The odd radius “2π cm” exists purely so that 2R = 4π × 10⁻² cancels the 4π in μ₀. Whenever you see a radius given as π or 2π cm, expect that cancellation and skip straight to it.

Option 7 × 10⁻⁵ is the trap for adding 3 + 4 directly.

1Given

• A conducting circular loop in a uniform magnetic field.
• Its plane is perpendicular to the field, so the field passes straight through it.
• Four proposed actions: translate it, rotate about a diameter, rotate about its own axis, deform it.

This one belongs to Chapter 6 (Electromagnetic Induction), not Chapter 4 — it uses Faraday's law, not any field formula.

2Asked

In which case an emf is induced.

3Concept & formula

An emf appears only when the flux changes. Three things can change it: B, A, or the angle θ.

4Steps

1. Starting position: plane perpendicular to the field means the loop's normal is along the field, so θ = 0 and Φ = BA, the maximum.
2. B is uniform and constant here, so B can never be the thing that changes. Only A and θ are candidates.
3. (a) Translated parallel to itself: the field is the same everywhere, so sliding the loop changes nothing — same B, same A, same θ. Φ is constant. No emf.
4. (b) Rotated about a diameter: the loop tips over, so θ changes with time, and Φ = BA cos θ changes. Emf induced. This is exactly how an AC generator works.
5. (c) Rotated about its own axis, parallel to the field: like spinning a coin flat-on. The plane stays perpendicular, the area is unchanged, θ stays 0. The loop just slides around within itself. No emf.
6. (d) Deformed from its original shape: the perimeter (length of wire) is fixed, and among all shapes of a given perimeter the circle has the largest area. So any deformation reduces A, and Φ = BA falls. Emf induced.
7. Both (b) and (d) induce an emf. WB JEE 2015 was a multiple-correct paper; single-key books print (b), the standard textbook case.

5Shortcut

Ask one question of any EMI problem: does B, A or θ change? If none of the three moves, the answer is zero.

Motion alone never induces an emf. Uniform field plus rigid translation is always zero.

1Given

• Solenoid length 0.4 (printed as cm — see the note below).
• Radius 1 cm.
• 400 turns of wire.
• Current 5 A.

2Asked

The magnetic field inside the solenoid.

3Concept & formula

Inside a long solenoid the field is uniform. Note what is absent from the formula: the radius. It plays no part, so “radius 1 cm” is a decoy.

4Steps

1. Find n, the turns per metre: n = N/L = 400 / 0.4.
2. 400 ÷ 0.4 = 1000, so n = 1000 turns per metre.
3. Substitute: B = (4π × 10⁻⁷)(1000)(5).
4. Multiply the plain numbers first: 1000 × 5 = 5000 = 5 × 10³.
5. So B = 4π × 10⁻⁷ × 5 × 10³ = 20π × 10⁻⁴.
6. 20π = 62.83, so B = 62.83 × 10⁻⁴.
7. Shift the decimal: B = 6.28 × 10⁻³ T.

On the printed units. Taken literally, 0.4 cm = 0.004 m would give n = 100000 and B = 0.628 T, which is not among the options — and a solenoid 0.4 cm long with a 1 cm radius would be wider than it is long, so the long-solenoid formula would not even apply. “cm” is a misprint for “m”. When the literal reading gives an answer that is absent, check the units before doubting your method.

5Shortcut

B = μ₀nI inside, and zero outside. At the very end of the solenoid the field halves: B(end) = μ₀nI/2. With an iron core, replace μ₀ by μ₀μᵣ.

1Given

• A toroid (a doughnut-shaped coil) with 200 turns.
• Current 1 A.
• Average radius 10 cm.

2Asked

The field at any point in the open space inside the toroid — that is, the empty hole in the middle, not the core.

3Concept & formula

A toroid has three distinct regions, and only one of them has a field.

4Steps

1. Read the region carefully: the question says the open space inside, which is the hole, not the core.
2. Draw a circular Amperian loop of radius r inside that hole.
3. Ask the only question that matters: how much current does it enclose? Every turn of the winding lies outside that loop.
4. So I(enclosed) = 0.
5. Ampère's law: B × 2πr = μ₀ × 0 = 0.
6. Hence B = 0.
7. The numbers 200, 1 A and 10 cm are there to let you compute the field in the core, which would be B = μ₀NI/2πr = (4π×10⁻⁷)(200)(1) / (2π × 0.1) = 4 × 10⁻⁴ T. That answers a different question, and its variants are planted among the options.

5Shortcut

Whenever a question hands you complete numerical data but asks about a region where B = 0, the data is bait. Read the region before touching the numbers.

A toroid confines its field entirely within the windings — that is why toroidal transformers are used where stray fields matter. Unlike a solenoid, the field in the core is not uniform; it falls as 1/r across the thickness.

1Given

• Experiment 1: a proton released from rest accelerates at a₀ toward the west.
• Experiment 2: the same proton projected north with speed v₀ accelerates at 3a₀ toward the west.

2Asked

The electric and magnetic fields in the room, with directions.

3Concept & formula

The Lorentz force has two parts. The electric part acts regardless of motion; the magnetic part needs motion. That is what lets you separate them.

4Steps

1. Take experiment 1 first. The proton is at rest, so v = 0, and the magnetic force qvB is zero. A magnetic field can do nothing to a stationary charge.
2. So the entire force is electric: eE = ma₀.
3. Hence E = ma₀ / e.
4. Direction of E: the proton is positive, so the force on it is along E. The force is west, so E is west. This alone eliminates the two “east” options.
5. Now experiment 2. The electric field has not changed, so it still supplies ma₀ west, whatever the proton is doing.
6. Total force is 3ma₀ west. Subtract the electric part: the magnetic force must be 3ma₀ − ma₀ = 2ma₀, also west.
7. The velocity is north and this force is west — perpendicular, so sin θ = 1 and the magnitude is simply ev₀B.
8. Set ev₀B = 2ma₀, giving B = 2ma₀ / ev₀.
9. Direction of B: use F = qv × B with the right hand. Fingers along v (north), curl toward B, thumb must give F (west).
10. Test B upward: north × up = east. Wrong way.
11. Test B downward: north × down = west. Correct.
12. So B = 2ma₀/ev₀, directed downward.

5Shortcut

Always solve the at rest case first — it hands you E for free, and the moving case then reduces to a subtraction.

Compass cross products worth memorising: north × up = east, north × down = west, east × up = south, north × east = up.

1Given

• Circular loop of radius R = 3 cm.
• A point on the axis at x = 4 cm from the centre.
• Field at that point = 54 μT.

2Asked

The field at the centre of the loop.

3Concept & formula

Both formulas share μ₀ and I, so divide them and those unknowns disappear. You never need to find the current.

4Steps

1. Write the ratio: B(centre) / B(axis) = [μ₀I / 2R] ÷ [μ₀IR² / 2(R²+x²)^(3/2)].
2. Dividing by a fraction means multiplying by its reciprocal: = (μ₀I / 2R) × [2(R²+x²)^(3/2) / μ₀IR²].
3. μ₀, I and the 2 all cancel: = (R²+x²)^(3/2) / R³.
4. Substitute the numbers: R² + x² = 3² + 4² = 9 + 16 = 25. (A 3-4-5 triangle — a designed number.)
5. (25)^(3/2) means √25 cubed: √25 = 5, and 5³ = 125.
6. R³ = 3³ = 27.
7. So the ratio is 125/27.
8. Therefore B(centre) = 54 × 125/27.
9. Simplify first: 54 ÷ 27 = 2.
10. So B(centre) = 2 × 125 = 250 μT.

5Shortcut

Memorise the combined form and save a minute: B(centre)/B(axis) = [(R²+x²)/R²]^(3/2).

The 54 was chosen to be divisible by 27 — a sign you are on the intended path.

1Given

• Two identical long wires AOB and COD, at right angles to each other, one above the other, crossing at O.
• Currents I₁ and I₂.
• Point P at distance d from O, along a direction perpendicular to the plane containing the wires.

2Asked

The magnetic field at P.

3Concept & formula

P is equidistant from both wires and sits on the perpendicular from each. Each field is perpendicular to its own wire, so the two fields are perpendicular to each other.

4Steps

1. P lies on the perpendicular through O, so its distance from each wire is d.
2. Field from wire 1: B₁ = μ₀I₁ / 2πd.
3. Field from wire 2: B₂ = μ₀I₂ / 2πd.
4. Each field is perpendicular to its own wire. The wires are at 90° to each other, so the two fields are also at 90°.
5. Perpendicular vectors combine by Pythagoras: B = √(B₁² + B₂²).
6. Factor out the shared μ₀/2πd: B = (μ₀ / 2πd) √(I₁² + I₂²).
7. Written with a fractional power: B = (μ₀ / 2πd)(I₁² + I₂²)^(1/2).

5Shortcut

Same rule as Q18: perpendicular sources never add arithmetically. The plain sum (I₁ + I₂) is always planted as a distractor.

If the two currents were equal, the answer would be μ₀I√2 / 2πd.

1Given

• A proton of mass m and charge q.
• Moving in a plane with kinetic energy E.
• Uniform magnetic field B, perpendicular to the plane of motion.

2Asked

The radius of the circular path.

3Concept & formula

The standard radius formula needs a speed; the question gives an energy. Convert one to the other.

4Steps

1. Start with r = mv / qB.
2. From E = ½mv², rearrange: v² = 2E/m, so v = √(2E/m).
3. Substitute: r = (m / qB) × √(2E/m).
4. Bring the leading m inside the root as m²: r = (1/qB) × √(m² × 2E/m).
5. Inside the root, m²/m = m, leaving 2Em.
6. So r = √(2Em) / qB.
7. Quicker route: momentum p = mv, and E = p²/2m gives p = √(2mE). Then r = p/qB = √(2mE)/qB in one line.

5Shortcut

Eliminate by units: any option with the charge q inside the square root cannot give metres.

The four faces of one formula — pick by what you are given:

1Given

• Galvanometer coil resistance G = 100 Ω.
• Permissible (maximum safe) current through the coil, Iɡ = 10 mA = 0.01 A.
• Desired ammeter range I = 1 A.

2Asked

The value of the resistance to be connected, and how.

3Concept & formula

The coil burns out above 10 mA, but the meter must read 1 A. Put a small resistance in parallel so most of the current bypasses the coil. Parallel branches share the same voltage.

4Steps

1. Convert the coil current to amperes: 10 mA = 10 × 10⁻³ = 0.01 A.
2. Current through the galvanometer branch: Iɡ = 0.01 A.
3. Current through the shunt branch: everything else, I − Iɡ = 1 − 0.01 = 0.99 A.
4. Both branches lie between the same two junctions, so the potential difference across them is identical.
5. Write that out: Iɡ G = (I − Iɡ) S, i.e. (0.01)(100) = (0.99) S.
6. Left side: 0.01 × 100 = 1.
7. So 1 = 0.99 S, giving S = 1 / 0.99.
8. 1 ÷ 0.99 = 1.0101…, which rounds to 1.01 Ω.

5Shortcut

Iɡ is tiny next to I, so the denominator is nearly 1 and S ≈ IɡG/I = 1 Ω. The answer must be just above 1 Ω — that alone picks 1.01 out of 1.9, 2.1 and 0.9.

Companion formula: for a voltmeter, put a large resistance in series, R = V/Iɡ − G. Ammeter in series with the circuit needs tiny resistance; voltmeter in parallel needs huge resistance.

1Given

• Solenoid 1: 200 turns per cm, current i, field at centre 6.28 × 10⁻² Wb/m².
• Solenoid 2: 100 turns per cm, current i/3.

2Asked

The field at the centre of the second solenoid.

3Concept & formula

4Steps

1. B depends on the product n × I, so form the ratio of the two solenoids.
2. Turns ratio: n₂/n₁ = 100/200 = 1/2.
3. Current ratio: i₂/i₁ = (i/3)/i = 1/3.
4. Multiply them: B₂/B₁ = (1/2)(1/3) = 1/6.
5. So B₂ = B₁ / 6 = 6.28 × 10⁻² / 6.
6. 6.28 ÷ 6 = 1.0466…, so B₂ = 1.0466 × 10⁻² ≈ 1.05 × 10⁻² Wb/m².

5Shortcut

Because it is a ratio, the “per cm” never has to be converted to “per metre” — the units cancel. Never convert inside a ratio unless the two sides use different units.

1Given

• A non-conducting disc of radius R.
• Rotating about the axis through its centre, perpendicular to its plane, with angular velocity ω.
• Charge q spread uniformly over its surface.

2Asked

The magnetic moment of the disc.

3Concept & formula

A disc is not one loop — the charge sits on an area, and the rings at different radii enclose different areas. So split it into rings and integrate.

4Steps

1. Surface charge density: σ = total charge / total area = q / πR².
2. Take a thin ring at radius x, of thickness dx. Its area is circumference × thickness = 2πx dx.
3. Charge on that ring: dq = σ(2πx dx).
4. That ring spins with the disc, so it goes round ω/2π times per second. Its current is dI = dq × ω/2π.
5. Substitute dq: dI = σ(2πx dx)(ω/2π) = σ ω x dx. (The 2π cancels.)
6. Magnetic moment of a loop is M = I × area. The ring's enclosed area is πx².
7. So dM = (σ ω x dx)(πx²) = σ ω π x³ dx.
8. Integrate x³ from 0 to R: ∫x³dx = R⁴/4.
9. So M = σ ω π R⁴ / 4.
10. Put σ = q/πR²: M = (q/πR²) × ω π R⁴ / 4.
11. π cancels, and R⁴/R² = R²: M = q ω R² / 4.

5Shortcut

Learn the pair: ring gives ½qωR², disc gives ¼qωR². The disc is exactly half the ring, because charge near the centre encloses less area.

Units check kills two options instantly: qωR² has units C·s⁻¹·m² = A·m², which is correct for a magnetic moment. Anything linear in R is dimensionally wrong.

1Given

• A rectangular loop carrying current i.
• A long straight wire carrying current I, parallel to one side of the loop and in the plane of the loop.
• From the figure, the near side's current runs the same way as I.

2Asked

Whether the loop rotates, moves away, moves toward the wire, or stays put.

3Concept & formula

4Steps

1. The top and bottom sides are mirror images about the mid-line, carrying current in opposite directions through the same field. Their forces cancel. Delete them.
2. Near side: its current is parallel to I, so it is attracted. It also sits where the wire's field is strongest.
3. Far side: its current is antiparallel to I, so it is repelled. But it sits further away, where the field is weaker.
4. Write the two magnitudes: F(near) ∝ 1/d₁ and F(far) ∝ 1/d₂, with d₁ < d₂.
5. Since d₁ is smaller, 1/d₁ is larger, so the attraction wins.
6. Net force is toward the wire. The loop moves bodily toward the wire.
7. Can it rotate? No. The wire lies in the loop's plane, so the field is perpendicular to the plane everywhere on the loop, and the torque about any axis is zero.

5Shortcut

A loop near a straight wire is always pulled in if its near side runs parallel to the wire's current — and always pushed away if it runs opposite. There is no third possibility; the near side always sits in the stronger field.

1Given

• Circular coil, radius 4 cm = 0.04 m, with 50 turns.
• Current 2 A.
• Magnetic field 0.1 Wb/m² = 0.1 T.
• Rotated through 180° from its equilibrium position.

2Asked

The work done in that rotation.

3Concept & formula

Work is the change in potential energy, not force times distance here.

4Steps

1. Convert the radius to metres first: 4 cm = 0.04 m.
2. Area of the coil: A = πr² = π(0.04)² = π(0.0016) = 1.6π × 10⁻³ m².
3. Magnetic moment: M = NiA = 50 × 2 × 1.6π × 10⁻³.
4. 50 × 2 = 100, so M = 100 × 1.6π × 10⁻³ = 0.16π ≈ 0.5027 A·m².
5. Identify the angles. “Equilibrium position” means the coil's moment is aligned with the field, so θ₁ = 0.
6. Rotating through 180° gives θ₂ = 180°.
7. cos 0° = 1 and cos 180° = −1, so cos θ₁ − cos θ₂ = 1 − (−1) = 2.
8. Therefore W = MB × 2 = 2MB.
9. Substitute: W = 2 × 0.5027 × 0.1 = 0.1005 J.
10. Rounded, W ≈ 0.1 J.

5Shortcut

Three standard turns: 0° → 90° costs MB; 0° → 180° costs 2MB; 90° → 180° costs MB. Read the starting angle carefully — “from equilibrium” always means θ₁ = 0.

Leaving the radius in centimetres inside πr² is the most common slip here, and it is off by a factor of 10⁴.

1Given

• Circular coil of radius 5 cm = 0.05 m.
• Current 0.9 A.
• ε₀ = absolute permittivity of air; c = 3 × 10⁸ m/s.

2Asked

The field at the centre, expressed using ε₀ rather than μ₀.

3Concept & formula

The link between magnetism and electrostatics is the speed of light.

4Steps

1. Ordinary calculation first: B = μ₀I / 2R = μ₀(0.9) / (2 × 0.05).
2. Denominator: 2 × 0.05 = 0.1.
3. 0.9 ÷ 0.1 = 9, so B = 9μ₀.
4. Now replace μ₀. From c = 1/√(μ₀ε₀), square both sides: c² = 1/(μ₀ε₀).
5. Rearrange: μ₀ = 1 / (ε₀c²).
6. Substitute: B = 9 / (ε₀c²).
7. Put in c = 3 × 10⁸, so c² = 9 × 10¹⁶.
8. B = 9 / (ε₀ × 9 × 10¹⁶).
9. The 9 on top cancels the 9 from c²: B = 1 / (ε₀ × 10¹⁶).

5Shortcut

The current 0.9 A was chosen so that its 9 cancels the 9 in c². Recognising the engineered number tells you the answer is the clean 1/(ε₀10¹⁶) without any messy arithmetic.

Keep the identity ready: μ₀ε₀c² = 1.

1Given

• Coil A: radius 25 cm = 0.25 m, current 10 A, 24 turns.
• Coil B: radius 15 cm = 0.15 m, current 15 A, 18 turns.
• Concentric, with the currents in opposite order.

2Asked

The magnetic induction at the common centre.

3Concept & formula

4Steps

1. Coil A: B(A) = μ₀(24)(10) / [2(0.25)].
2. Numerator: 24 × 10 = 240. Denominator: 2 × 0.25 = 0.5.
3. 240 ÷ 0.5 = 480, so B(A) = 480μ₀.
4. Coil B: B(B) = μ₀(18)(15) / [2(0.15)].
5. Numerator: 18 × 15 = 270. Denominator: 2 × 0.15 = 0.3.
6. 270 ÷ 0.3 = 900, so B(B) = 900μ₀.
7. Both coils are concentric and coplanar, so their fields lie along the same line. The currents run in opposite senses, so the fields oppose.
8. Subtract: B = 900μ₀ − 480μ₀ = 420μ₀ T.
9. Direction: that of the stronger coil, B.

5Shortcut

Do the subtraction before you reach for μ₀ — the constant just rides along, and the options are written in μ₀ anyway.

Sanity check on the options: 480μ₀ is there for anyone who computes only one coil, and there is no option at 1380μ₀, which confirms subtraction was intended.

1Given

• A power line carrying current from west to east.
• The point of interest is a short distance below the wire.

2Asked

The direction of the magnetic field there.

3Concept & formula

The field circles the wire. Right-hand grip rule: thumb along the current, fingers curl the way the field goes.

4Steps

1. Point the right thumb east — the direction of the current.
2. Curl the fingers. They wrap around the wire in closed circles.
3. Above the wire the fingers run one way; below the wire they run the opposite way. So you must be careful which side is asked for.
4. Check with the cross product. The direction of B goes as Î × r̂, where r̂ points from the wire to the field point.
5. Here Î = east and r̂ = down (the point is below).
6. east × down = north.
7. So the field below the wire points from south to north.
8. (Above the wire it would point north to south — the exact reverse.)

5Shortcut

For a horizontal wire running east-west, the field below and the field above are always opposite. Read the word “below” or “above” before answering — that single word decides between two of the four options.

1Given

• Two wires of the same material and same length, 20 cm each.
• Coil A has 4 turns; coil B has 2 turns.
• The same current flows in both.

2Asked

The ratio of the magnetic induction at the centres of A and B.

3Concept & formula

The length is fixed, so more turns forces a smaller radius. N therefore enters twice.

4Steps

1. The wire of length L is wound into N circles, so L = N × 2πR.
2. Rearrange for the radius: R = L / 2πN.
3. Field at the centre of an N-turn coil: B = μ₀NI / 2R.
4. Substitute R: B = μ₀NI / [2 × L/(2πN)].
5. The denominator is 2L/2πN = L/πN.
6. Dividing by L/πN means multiplying by πN/L: B = μ₀NI × πN / L.
7. So B = πμ₀N²I / L, which shows B ∝ N².
8. Ratio: B(A)/B(B) = (4/2)² = 2² = 4.
9. So the ratio is 4 : 1. Note the 20 cm never entered — it is common to both and cancels.

5Shortcut

Same as Set 1 Q1 and Q9 in different clothes. The trigger words here are “same material and same length”. Whenever the wire is shared, B ∝ N².

1Given

• A long straight current-carrying wire.

2Asked

The symmetry of the magnetic field around it.

3Concept & formula

4Steps

1. The field lines are closed circles wrapped around the wire.
2. Look at what B depends on: only the perpendicular distance r.
3. Move along the wire — B is unchanged. Move around the wire at fixed r — B is unchanged in magnitude.
4. “Same everywhere at the same distance from an axis” is precisely cylindrical symmetry.
5. Spherical symmetry would mean the field depended only on distance from a point, which is false here.
6. This symmetry is exactly what lets you pull B outside the integral in Ampère's law: ∮B·dl = B(2πr) = μ₀I.

5Shortcut

Match the symmetry to the source: a point charge gives spherical, a long wire gives cylindrical, a large sheet gives planar. Ampère's and Gauss's laws are only usable when one of these applies.

1Given

• Current I flowing along an infinitely long straight thin-walled tube.
• A point inside the tube, at distance r from the wall.

“Thin walled” means all the current travels on the surface — none of it is spread through the interior.

2Asked

The magnetic induction at that interior point.

3Concept & formula

4Steps

1. Draw an Amperian circle passing through the point, centred on the tube's axis and lying inside the hollow.
2. Ask the standard question: how much current does this circle enclose?
3. All the current flows on the wall, which lies outside the circle. So I(enclosed) = 0.
4. Ampère's law gives B × 2πr = μ₀ × 0 = 0.
5. Since 2πr is not zero, B must be zero.
6. This holds at every point inside, whatever the distance from the wall — which is why the distance r in the question is irrelevant.

5Shortcut

One word decides this question. Compare with Set 1 Q4:

1Given

• Assertion: in a moving-coil galvanometer the torque on the coil is maximum in any position of the coil.
• Reason: the concave magnetic poles make the field radial, so the coil stays correctly oriented to the field lines even after rotating.

2Asked

Whether each statement is true, and whether the reason explains the assertion.

3Concept & formula

4Steps

1. General torque on a coil in a field: τ = NIAB sin θ.
2. In an ordinary uniform field, θ changes as the coil turns, so τ falls as it rotates — and the deflection would not be proportional to the current.
3. Now add curved (concave) pole pieces with a soft-iron cylinder between them. This makes the field radial, pointing outward from the axis at every position.
4. With a radial field, the coil's plane always contains the field lines, whatever angle it has turned to. So θ stays at 90°.
5. sin 90° = 1, so τ = NIAB, its maximum value, in every position. The assertion is true.
6. The reason describes exactly this arrangement, so it is true as well.
7. Does the reason explain the assertion? Yes — the radial field is precisely why sin θ never drops below 1. So R is the correct explanation of A.
8. The purpose: balancing torque against the spring gives NIAB = kφ, hence I ∝ φ — a linear scale. That is the whole reason the design exists.

Wording note: the reason says the plane of the coil is always perpendicular to the lines of induction. Strictly, the plane is always parallel to the field, and it is the coil's normal that is perpendicular. Textbooks print it both ways; the intended meaning is the radial-field arrangement, and the marked key is unaffected.

5Shortcut

For assertion–reason questions, judge A and R separately first, and only then ask whether R explains A. Trying to judge them together is where marks get lost.

1Given

• A charged particle in circular motion in a uniform magnetic field.
• Four candidates: speed, mass, charge, magnetic induction.

2Asked

Which of them the time period does not depend on.

3Concept & formula

4Steps

1. Derive it once. From qvB = mv²/r, cancel one v to get r = mv/qB.
2. One lap takes T = 2πr / v.
3. Substitute r: T = (2π/v)(mv/qB).
4. The v cancels, leaving T = 2πm / qB.
5. Now simply read the formula. It contains m, q and B — so the period depends on all three.
6. It does not contain v or r. So the period is independent of speed.
7. Physically: a faster particle takes a bigger circle, and the extra distance exactly offsets the extra speed.

5Shortcut

“Independent of” questions are answered by looking at what is absent from the formula. Write the formula, then read off what is missing.

This speed-independence is the entire operating principle of the cyclotron.

1Given

• A positively charged particle.
• Magnetic field B î (along +x).
• Velocity v ĵ (along +y).

2Asked

The instantaneous direction associated with the particle.

3Concept & formula

4Steps

1. Write the force: F = q(vĵ) × (Bî).
2. Pull the scalars out: F = qvB (ĵ × î).
3. Now the key step — the order. The cyclic rule is î × ĵ = k̂. Reversing the order flips the sign, so ĵ × î = −k̂.
4. Therefore F = −qvB k̂, which points along the negative z-axis.
5. So the force is along −Z, and the axis involved is the Z-axis.
6. The particle's own instantaneous velocity is of course along +Y, so the wording is ambiguous; the marked key is the Z-axis, which is where the force acts and where the motion is being turned.

5Shortcut

Memorise the cycle î → ĵ → k̂ → î. Going forwards along the cycle gives a plus sign, going backwards gives a minus. So k̂ × î = +ĵ but î × k̂ = −ĵ.

And always check the sign of the charge: for an electron the force is opposite to v × B.

1Given

• Two points A and B on the axis of a circular loop.
• Distances from the centre: x(A) = 4 cm and x(B) = 3√3 cm.
• Ratio of the fields at A and B is 216 : 125.

2Asked

The radius of the loop.

3Concept & formula

4Steps

1. For one fixed loop, μ₀, I and R² are the same at both points, so only the bracket varies: B ∝ 1/(R²+x²)^(3/2).
2. Because it sits in the denominator, the bigger field belongs to the smaller bracket. Write the ratio the right way up: B(A)/B(B) = [(R²+x(B)²) / (R²+x(A)²)]^(3/2).
3. Set that equal to 216/125.
4. Notice 216 = 6³ and 125 = 5³. So take the cube root of both sides first.
5. Raising to the power 1/3 turns the exponent 3/2 into 1/2: [(R²+x(B)²)/(R²+x(A)²)]^(1/2) = 6/5.
6. Square both sides: (R²+x(B)²)/(R²+x(A)²) = 36/25.
7. Compute the squares: x(A)² = 4² = 16, and x(B)² = (3√3)² = 9 × 3 = 27.
8. So (R² + 27)/(R² + 16) = 36/25.
9. Cross-multiply: 25(R² + 27) = 36(R² + 16).
10. Expand: 25R² + 675 = 36R² + 576.
11. Collect: 675 − 576 = 36R² − 25R², so 99 = 11R².
12. R² = 9, hence R = 3 cm.

5Shortcut

Spot the perfect cubes. Any ratio like 216:125, 27:8 or 343:216 in an axial-field question is there so you can cube-root it and kill the 3/2 power in one move.

Watch the direction of the ratio: B is inversely related to the bracket, so the bracket ratio is the reciprocal of the field ratio.

1Given

• Statement I: Biot–Savart's law gives the magnetic field of an infinitesimal current element (Idl).
• Statement II: it is analogous to Coulomb's inverse-square law, with the field produced by a scalar source Idl, while Coulomb's field comes from a vector source q.

2Asked

Which of the two statements is correct.

3Concept & formula

4Steps

1. Statement I is simply the definition of the law, and it is correct as written. True.
2. Now check Statement II piece by piece.
3. Is Biot–Savart analogous to Coulomb, with an inverse-square dependence? Yes — both fall off as 1/r².
4. Is the current element Idl a scalar? No. dl is a length vector pointing along the current, which is why the formula contains a cross product.
5. Is charge q a vector? No. Charge is a plain scalar.
6. So Statement II has the two sources swapped. It is incorrect.
7. Correct version: Coulomb's source is scalar (q); Biot–Savart's source is vector (Idl).

5Shortcut

Remember the contrast as a pair:

The presence of a cross product in the formula is itself the proof that the source is a vector.

1Given

• Coils X and Y have equal turns and carry equal currents in the same sense.
• Both subtend the same solid angle at the point O.
• The smaller coil X is midway between O and Y.

2Asked

The ratio B(Y) / B(X) at the point O.

3Concept & formula

Equal solid angle means the two coils fit the same cone, so radius and distance scale together.

4Steps

1. X is midway between O and Y, so if X sits at distance d, then Y sits at 2d.
2. Same solid angle means the ratio radius ÷ distance is the same for both: R(X)/d = R(Y)/2d.
3. Cancel d: R(Y) = 2R(X). So going from X to Y, both the radius and the distance double.
4. Call the scale factor k = 2, so every length in the formula gets multiplied by k.
5. Look at the formula's structure. The numerator has R², so it picks up k².
6. The denominator has (R² + x²)^(3/2). Multiplying every length by k multiplies R² + x² by k², and then the 3/2 power makes it (k²)^(3/2) = k³.
7. So the whole field scales as k²/k³ = 1/k.
8. With k = 2, B(Y) = B(X)/2.
9. Hence B(Y)/B(X) = 1/2.

5Shortcut

Once N and I are fixed, B has the dimensions of 1/length. So if the entire geometry is scaled up by k, the field must scale by 1/k — no algebra needed at all.

1Given

• Two very long parallel wires, each carrying current i, in opposite directions.
• Separation d.
• A point charge q sits equidistant from both wires, in the plane of the wires.
• Its velocity v is perpendicular to that plane.

2Asked

The magnitude of the magnetic force on the charge at that instant.

3Concept & formula

4Steps

1. Find the field first. The charge is midway, so it is d/2 from each wire.
2. Each wire gives μ₀i / [2π(d/2)] = μ₀i / πd.
3. The currents are opposite, so at the midpoint the two fields point the same way and add: B = 2μ₀i / πd. (Note: if the currents were parallel, they would cancel here instead.)
4. Now the crucial step — the direction of that field. At any point in the plane of the wires, the field of a straight wire points perpendicular to that plane.
5. The velocity is also stated to be perpendicular to the same plane.
6. So v and B lie along the same line: the angle between them is 0°.
7. F = qvB sin 0° = 0.
8. The field is large and the charge is moving, and yet the force is exactly zero.

5Shortcut

Three of the four options are variations on the magnitude of B — the trap is to compute B and stop. Always check the angle between v and B before computing a force.

A magnetic field exerts no force on a charge moving along it, no matter how strong the field is.

1Given

• Two wires of the same length L.
• One is shaped into a square, the other into a circle.
• Both carry the same current.

2Asked

The ratio of their magnetic moments.

3Concept & formula

Same current and one turn each, so M = IA depends only on the enclosed area.

4Steps

1. Square: the perimeter is the whole wire, so 4 × side = L, giving side = L/4.
2. Its area: A(sq) = (L/4)² = L²/16.
3. Circle: the circumference is the whole wire, so 2πr = L, giving r = L/2π.
4. Its area: A(cir) = πr² = π(L/2π)² = πL²/4π² = L²/4π.
5. Take the ratio: A(sq)/A(cir) = (L²/16) ÷ (L²/4π).
6. Dividing by a fraction means multiplying by its reciprocal: = (L²/16)(4π/L²).
7. L² cancels: = 4π/16 = π/4.
8. So M(square) : M(circle) = π : 4.

5Shortcut

Since π ≈ 3.14, the ratio π/4 is less than 1 — the circle has the bigger moment. That is the general rule: for a fixed perimeter, the circle encloses the largest area of any shape.

The same fact answered Set 1 Q19, where deforming a circular loop always reduces its flux.

1Given

• A wire of length l bent into a circular coil.
• Current I established in it, placed in a uniform field B.
• The maximum torque is stated to be I B l² / (xπ).

2Asked

The value of x.

3Concept & formula

4Steps

1. The wire of length l forms one circle, so its circumference is l: 2πr = l.
2. Rearrange for the radius: r = l / 2π.
3. Area of the coil: A = πr² = π(l/2π)².
4. Expand the square: (l/2π)² = l²/4π².
5. So A = π × l²/4π² = l²/4π. (One π cancels.)
6. Magnetic moment: M = IA = I l²/4π.
7. Torque is maximum when sin θ = 1, so τ(max) = MB = I B l² / 4π.
8. Compare with the given form I B l²/(xπ): the denominators must match, so xπ = 4π.
9. Therefore x = 4.

The question says “of some turns”. With n turns the radius becomes l/2πn and the torque works out to I B l²/4πn, so x = 4n. Since the printed answer contains no n, a single turn is intended — and 4 is the option.

5Shortcut

Worth memorising outright: a wire of length l bent into one circle encloses area l²/4π. It shows up in torque, moment and flux questions alike.

Compare with a square from the same wire: area l²/16, which is smaller — consistent with Q19.

1Given

• An electron (negative charge) moving along the positive x-axis.
• It must be made to travel on an anticlockwise circular path in the xy-plane.

2Asked

The direction of the magnetic field to be applied.

3Concept & formula

4Steps

1. Work out which way the force must point. The electron moves along +x.
2. Anticlockwise in the xy-plane means it turns to the left, curving up toward +y.
3. The centripetal force always points toward the centre of the circle, so the force must be along +y.
4. Now test a field along +z. Write F = (−e)(v î) × (B k̂).
5. Pull the scalars out: F = −evB (î × k̂).
6. Use the cycle: k̂ × î = +ĵ, so reversing gives î × k̂ = −ĵ.
7. Substitute: F = −evB(−ĵ) = +evB ĵ.
8. The force is along +y — exactly what was needed. So the field is along the positive z-axis.
9. Sanity check: a positive charge moving along +x in a +z field would feel a force along −y and would go clockwise. The electron, being negative, goes the opposite way — anticlockwise. Consistent.

5Shortcut

Solve it for a positive charge first, then flip the answer for an electron. Two sign reversals to track: the charge, and the order of the cross product.

Rule of thumb in the xy-plane: a positive charge in a +z field circles clockwise; a negative charge circles anticlockwise.

1Given

• A charged particle projected into a transverse uniform magnetic field.
• Four proposed statements about the area enclosed by its path: (i) ∝ kinetic energy, (ii) ∝ 1/charge², (iii) ∝ velocity, (iv) ∝ 1/time.

2Asked

Which of the statements are correct.

3Concept & formula

4Steps

1. Start from r = mv/qB and square it: r² = m²v²/q²B².
2. Area of the circle: A = πr² = πm²v² / q²B².
3. Test (i). Replace v² using E = ½mv², which gives v² = 2E/m.
4. Substitute: A = (πm²/q²B²)(2E/m) = 2πmE / q²B².
5. E appears to the first power on top, so A ∝ E. Statement (i) is correct.
6. Test (ii). Look at the denominator: q is squared there, so A ∝ 1/q². Statement (ii) is correct.
7. Test (iii). The formula has v², not v. Doubling the speed makes the area four times bigger, not twice. Statement (iii) is wrong.
8. Test (iv). The time period is T = 2πm/qB, so qB = 2πm/T.
9. Substitute that into A = πm²v²/q²B²: A = πm²v² / (2πm/T)² = πm²v²T² / 4π²m² = v²T² / 4π.
10. So A goes as T², and directly, not inversely. Statement (iv) is wrong.
11. Only (i) and (ii) hold.

5Shortcut

Derive A = 2πmE/q²B² once and read all four claims straight off it. Every “proportional to” question is answered by looking at the powers in a single expression.

1Given

• A vertical circular coil of one turn, radius 9.42 cm = 0.0942 m.
• Its plane lies in the magnetic meridian.
• A short magnetic needle pivoted at the centre, free to rotate horizontally.
• Current 6 A; Earth's horizontal field B(H) = 4 × 10⁻⁵ T.

2Asked

The angle through which the needle deflects.

3Concept & formula

The magnetic meridian is the vertical plane containing Earth's horizontal field. With the coil in that plane, the coil's own field (perpendicular to its plane) is at right angles to B(H). The needle settles along the resultant.

4Steps

1. Compute the coil's field: B = μ₀NI / 2R = (4π × 10⁻⁷)(1)(6) / [2 × 0.0942].
2. Numerator: 4π × 6 = 24π, so it is 24π × 10⁻⁷ = 75.4 × 10⁻⁷ = 7.54 × 10⁻⁶.
3. Denominator: 2 × 0.0942 = 0.1884.
4. Divide: 7.54 × 10⁻⁶ / 0.1884 = 4.0 × 10⁻⁵ T.
5. Both fields are horizontal and mutually perpendicular, so the needle points along their resultant, at angle θ from B(H) with tan θ = B(coil)/B(H).
6. Substitute: tan θ = 4 × 10⁻⁵ / 4 × 10⁻⁵ = 1.
7. θ = tan⁻¹(1) = 45°.

5Shortcut

The radius 9.42 cm is 3π/100 m, chosen so that 2R = 6π/100 cancels the 6 and the π from the numerator, landing exactly on 4 × 10⁻⁵ — the same value as B(H). Equal fields always mean 45°.

The general relation is I = k tan θ, where k is called the reduction factor. That is why the instrument is called a tangent galvanometer.

1Given

• A magnetic pole of strength m = 8 A·m.
• Moved once around a long straight wire.
• The wire carries I = 5 A.

2Asked

The work done in taking the pole around the wire.

3Concept & formula

4Steps

1. The force on a magnetic pole of strength m in a field B is F = mB.
2. Work done along a path is W = ∮F·dl. Since m is a constant, pull it out: W = m ∮B·dl.
3. The path encircles the wire, so Ampère's law applies directly: ∮B·dl = μ₀I.
4. Therefore W = μ₀ m I.
5. Substitute: W = (4π × 10⁻⁷)(8)(5).
6. Multiply the plain numbers: 4 × 8 × 5 = 160. So W = 160π × 10⁻⁷.
7. Rewrite to match the options: 160 × 10⁻⁷ = 16 × 10⁻⁶.
8. So W = 16π × 10⁻⁶ J.

5Shortcut

Why isn't it zero? For a closed path you would normally expect zero work — true for gravity and for electrostatics, which are conservative. But the magnetic field of a steady current is non-conservative: ∮B·dl = μ₀I ≠ 0. Every lap around the wire extracts real work, and n laps give n times as much.

The option “zero” is planted for anyone applying the electrostatics habit.