Force on a Conductor & Force Between Parallel Currents
Priority 4 for the next ILTS. The push on a wire, the balance-against-gravity family that NEET has asked three times, and why two wires pull each other together.
Two facts you already know, joined together: a wire with current makes a magnetic field, and a wire in a magnetic field feels a push. Put them side by side and two wires start pushing each other.
1. Why a wire feels a push at all
A wire carrying current is just a crowd of moving charges walking along together. From the previous topic we know a magnet pushes each moving charge sideways. So if the magnet pushes every charge inside the wire, it must push the whole wire.
That single idea is why electric motors spin, why loudspeakers make sound, and why a train on a maglev track floats.
How the book gets there: if there are n free charges packed into every cubic metre, then a rod of length l and cross-section A holds nlA of them. Each drifts along slowly at speed vd. Multiply the force on one charge by how many there are, then bundle nqvdA back into the single familiar word current. Everything collapses to one neat line.
F = I l × B ⇒ F = B I L sin θ
This is the whole idea behind electric motors, loudspeakers and maglev trains: a magnet cannot grab a wire, but it can grab every charge inside it.
Longer wire, more current, stronger field → bigger shove. The angle θ is measured between the wire and the field.
A small but exam-worthy detail. Current itself is just a number, not an arrow. So the "direction" job is handed to the length l, which points whichever way the current flows.
2. When the push is zero
Same rule as for a single charge: if the wire lies along the field, θ = 0 and sin θ = 0, so there is no force at all. The field only pushes a wire that cuts across it.
NCERT's own example makes this vivid. The Earth's horizontal field runs south to north. A wire carrying current east to west cuts straight across it and feels a push. The same wire carrying current south to north lies along the field and feels nothing.
A wire lying along the field lines feels no force whatever the current. NCERT's own example: a wire carrying current south to north in the Earth's field feels nothing; turn it east to west and it feels the full push.
3. Which way does it push?
The force is perpendicular to both the wire and the field. Find it with the right hand: point your fingers along the current, curl them towards B, and your thumb gives the force. (Some books use Fleming's left-hand rule for this; either is fine as long as she uses one consistently.)
Flip the current and the push flips with it. The force is always perpendicular to both the wire and the field — never along either of them.
The most important warning in this topic. In F = I l × B, the field B must come from outside. A wire makes its own magnetic circles, but a wire cannot push itself — just as you cannot lift yourself by pulling your own shoelaces.
4. Bent wires and closed loops
What if the wire is curved? Chop it into tiny straight pieces, work out the push on each, and add them up. When you do that for a wire joining point A to point B, something lovely happens: only the straight-line distance from A to B matters, not the winding path in between.
F = I Leffective B sin θ, where Leffective = straight line joining the two ends
Picture it: a semicircular wire of radius R in a uniform field feels the same force as a straight wire of length 2R (the diameter) joining its two ends. The bendy path makes no difference to the total.
And that leads straight to the most-tested consequence:
The net force on any CLOSED loop in a uniform magnetic field is ZERO. A closed loop starts and finishes at the same point, so the straight-line distance between its ends is zero, so the effective length is zero. (There is still a torque — that was Priority 1 — but no net force.)
Watch the wire writhe about while the dashed line stays put. In a uniform field the shape is irrelevant — and a closed loop, whose ends coincide, feels no net force at all.
5. The balance family — NEET's favourite question here
Take a wire lying in a magnetic field. Gravity pulls it down with force mg. The magnetic force pushes it up with force BIL. If the wire hangs perfectly still in mid-air, the two must be exactly equal — a tug of war where neither side moves:
B I L = m g
Write down what pushes up and what pulls down, then set them equal. That single move solves the 2018 incline, the 2023 suspended wire and the 2026 levitation question.
This one equation, rearranged three different ways, is the whole family:
What the question asks for
Rearrange to
The field needed to hold the wire up
B = mg / IL
The current needed
I = mg / BL
The mass that can be supported
m = BIL / g
The incline version. Sometimes the wire rests on a slope at angle α, held from sliding by a vertical field. Then the component of gravity down the slope, mg sin α, is balanced by the component of the magnetic force along the slope, BIL cos α, giving:
I = (m/L) g tan α / B
Why this family matters so much. NEET asked it in 2018 (rod on an incline), 2023 (wire suspended in mid-air) and 2026 (wire held down by another wire). Three separate years, one method: write down what pushes up, write down what pulls down, set them equal.
6. Two wires talking to each other
Now join the two halves of the chapter. Wire A makes a magnetic field. Wire B sits inside that field and gets pushed. But wire B also makes a field, and wire A sits inside that and gets pushed back.
Picture it: two people in a swimming pool. Each one makes waves, and each one is rocked by the other's waves. Neither has to touch the other.
Wire A's field at wire B's position is Ba = μ₀Ia/2πd. Wire B, carrying Ib, then feels F = IbLBa. Putting those together:
F = μ₀ Ia Ib L / 2πd f = μ₀ Ia Ib / 2πd
Two people in a swimming pool: each makes waves, and each is rocked by the other's. Nothing here is new — it is the straight-wire field fed straight into the force on a wire.
The second form is force per metre of wire, and that is the version NEET uses, because it avoids having to say how long the wires are.
7. Attract or repel?
Parallel currents attract. Antiparallel currents repel.
Currents flowing the same way → the wires pull together.
Currents flowing opposite ways → the wires push apart.
This is the reverse of what you learnt about charges, and it catches almost everyone. Two like charges (both positive) repel. Two like currents (both the same direction) attract. Reasoning by analogy gives the wrong answer, so learn this pair as a fact.
Watch the wires actually move. Like currents attract — the exact reverse of like charges, which repel. Reasoning by analogy here gives the wrong answer every time.
8. Newton's third law shows up uninvited
The push on A from B and the push on B from A come out exactly equal and exactly opposite: Fab = −Fba. Newton wrote that law 150 years before anyone knew magnetism existed, yet these formulas obey it all by themselves. When two separate corners of physics agree without being asked, it is a sign both are right.
Newton wrote his third law 150 years before anyone knew magnetism existed, yet these formulas obey it on their own. When two separate corners of physics agree without being asked, it is a sign both are right.
9. Defining the ampere
Put I₁ = I₂ = 1 A and d = 1 m into the force-per-metre formula. Since μ₀/2π = 2 × 10⁻⁷, you get:
f = 2 × 10⁻⁷ N per metre
So one ampere is the steady current which, flowing in two infinitely long parallel wires one metre apart in vacuum, makes them attract with a force of 2 × 10⁻⁷ newtons per metre of length. The ampere is one of the seven SI base units, and this is the elegant part: we define electric current by something we can measure mechanically — a tiny pull between two wires.
Once the ampere is fixed, the coulomb follows: 1 coulomb is the charge that flows in 1 second when the current is 1 ampere.
The elegant part: electric current, an invisible thing, is pinned down by something you can measure mechanically — a tiny pull between two wires. And once the ampere is fixed, the coulomb follows.
10. A loop near a straight wire
Place a square loop flat beside a long straight wire. Four arms, four forces:
The two arms perpendicular to the wire feel equal and opposite forces that cancel.
The near arm and the far arm are both parallel to the wire, but at different distances.
Since the force falls off as 1/d, the near arm always wins. If the near arm's current runs the same way as the wire's, the loop is pulled towards the wire:
Fnet = (μ₀ I i / 2π) [ 1/x − 1/(x+a) ]
Note this does not contradict "net force on a closed loop is zero" — that rule needs a uniform field, and the field near a straight wire is anything but uniform.
A tug of war the near arm always wins, because the wire's field is stronger close in. Note this does not contradict "no net force on a closed loop" — that rule needs a uniform field.
Part 2 — Formula sheet
Everything in sections 4.2.3 and 4.8, on one page.
Quantity
Formula
Watch out for
Force on a straight wire
F = B I L sin θ
θ is between the wire and B.
Vector form
F = I l × B
Current is a scalar; l carries the direction.
Maximum force
F = B I L
When the wire is perpendicular to B.
Zero force
F = 0
When the wire lies along B (θ = 0° or 180°).
Force per unit length on a wire
f = B I sin θ
Answer in newtons per metre.
Bent wire in a uniform field
F = I Leff B sin θ
Leff = straight line joining the two ends.
Closed loop in a uniform field
Fnet = 0
Only in a uniform field. Torque may still exist.
Wire suspended in mid-air
B I L = m g
Convert grams to kilograms first.
Rod on an incline of angle α
I = (m/L) g tan α / B
Vertical field; balance along the slope.
Force between parallel wires
F = μ₀ I₁ I₂ L / 2πd
Total force on a length L.
Force per unit length
f = μ₀ I₁ I₂ / 2πd = (2×10⁻⁷) I₁I₂ / d
The form NEET uses. Falls as 1/d, not 1/d².
Nature of the force
Parallel → attract Antiparallel → repel
The opposite of the rule for charges.
Newton's third law
Fab = − Fba
Equal and opposite for steady currents.
Definition of the ampere
f = 2 × 10⁻⁷ N/m
Two wires, 1 A each, 1 m apart, in vacuum.
Definition of the coulomb
1 C = 1 A × 1 s
Follows once the ampere is fixed.
Levitation of one wire above another
μ₀I₁I₂ / 2πh = λ g
λ = mass per unit length. Currents must be parallel.
Loop near a straight wire
F = (μ₀ I i / 2π)[1/x − 1/(x+a)]
The near arm dominates; the field is non-uniform.
The three lookalike formulas — keep them apart. All three involve a straight wire, and the third is simply the first fed into the second.
Situation
Formula
Unit of the answer
Field at distance r from one wire
B = μ₀I / 2πr
tesla
Force on a wire in an external field
F = BIL sin θ
newton
Force per metre between two wires
f = μ₀I₁I₂ / 2πd
newton per metre
If she derives the third one rather than memorising it, she can never misplace the 2π.
Check the unit in the options first. Tesla means they want a field. Newton means a force on a wire. Newton per metre means the two-wire formula. That one glance rules out two thirds of the wrong answers before any arithmetic starts.
Part 3 — 50 questions with step-by-step solutions
Attempt each one on paper first, then open the solution. Every solution follows the same five steps: Given → Asked → Concept → Formula → Solution.
Q1Force on wire · numeric
A straight wire of length 1 m carrying a current of 2 A is placed perpendicular to a magnetic field of 0.5 T. The force on the wire is:
(a) 0.5 N
(b) 1 N
(c) 2 N
(d) 4 N
Show step-by-step solution
GivenL = 1 m, I = 2 A, B = 0.5 T, θ = 90°
AskedForce F on the wire
ConceptA wire perpendicular to the field feels the maximum force, since sin θ = 1.
FormulaF = B I L sin θ
Solutionsin 90° = 1 F = 0.5 × 2 × 1 × 1 F = 1 N
Answer: 1 N
Q2Force on wire · numeric
A wire of length 50 cm carries 5 A and is placed at 30° to a magnetic field of 0.2 T. The force on it is:
(a) 0.125 N
(b) 0.25 N
(c) 0.5 N
(d) 1 N
Show step-by-step solution
GivenL = 50 cm = 0.5 m, I = 5 A, B = 0.2 T, θ = 30°
AskedForce F
ConceptThe wire is not perpendicular, so the sin θ factor must be included. Convert cm to m first.
FormulaF = B I L sin θ
Solutionsin 30° = 0.5 F = 0.2 × 5 × 0.5 × 0.5 = 0.5 × 0.5 = 0.25 N
Answer: 0.25 N
Q3Force on wire · concept
A current-carrying wire is placed parallel to a uniform magnetic field. The force acting on it is:
(a) BIL
(b) BIL/2
(c) zero
(d) maximum
Show step-by-step solution
GivenWire parallel to B, so θ = 0°
AskedForce on the wire
ConceptThe sin θ factor vanishes when the wire lies along the field — the same rule as for a single moving charge.
FormulaF = B I L sin θ
Solutionθ = 0° between the wire and the field. sin 0° = 0 F = 0 — the field only pushes a wire that cuts across it.
Answer: zero
Q4Force on wire · numeric
A wire of length 2 m carries a current of 4 A perpendicular to a field. If it experiences a force of 8 N, the magnetic field is:
(a) 0.5 T
(b) 1 T
(c) 2 T
(d) 4 T
Show step-by-step solution
GivenL = 2 m, I = 4 A, F = 8 N, θ = 90°
AskedMagnetic field B
ConceptRearrange the force formula to make B the subject.
FormulaB = F / (I L sin θ)
SolutionB = 8 / (4 × 2 × 1) B = 8 / 8 = 1 T
Answer: 1 T
Q5Force on wire · Earth
The Earth's horizontal magnetic field is 3 × 10⁻⁵ T, directed from south to north. A long wire on a horizontal table carries 1 A from east to west. The force per unit length on it is:
(a) zero
(b) 3 × 10⁻⁵ N/m
(c) 1.5 × 10⁻⁵ N/m
(d) 6 × 10⁻⁵ N/m
Show step-by-step solution
GivenB = 3 × 10⁻⁵ T (south to north), I = 1 A (east to west)
AskedForce per unit length f
ConceptEast-to-west current cuts across a south-to-north field at right angles, so θ = 90°.
Formulaf = B I sin θ
SolutionThe current direction is perpendicular to the field direction ⇒ θ = 90°. f = 3 × 10⁻⁵ × 1 × 1 = 3 × 10⁻⁵ N/m (directed downwards)
Answer: 3 × 10⁻⁵ N/m
Q6Force on wire · Earth
The same wire now carries its 1 A current from south to north, along the Earth's horizontal field. The force per unit length is:
(a) 3 × 10⁻⁵ N/m
(b) 1.5 × 10⁻⁵ N/m
(c) zero
(d) 6 × 10⁻⁵ N/m
Show step-by-step solution
GivenB from south to north, I from south to north
AskedForce per unit length
ConceptNow the wire lies along the field, so there is no force at all.
Formulaf = B I sin θ
SolutionThe current is parallel to the field ⇒ θ = 0°. sin 0° = 0 f = 0
Answer: zero
Q7Force on wire · direction
A horizontal wire carries current towards the east in a magnetic field pointing towards the north. The force on the wire is directed:
(a) upwards
(b) downwards
(c) towards the west
(d) towards the south
Show step-by-step solution
GivenI towards east, B towards north
AskedDirection of the force
ConceptSet up axes and use the cross product rather than guessing.
FormulaF = I l × B; î × ĵ = k̂
SolutionTake x = east, y = north, z = up. l is along +x (east), B is along +y (north). l × B = î × ĵ = k̂, which is upwards. So the force is vertically upwards.
Answer: upwards
Q8Force on wire · effective length
A wire bent into a semicircle of radius R carries current I in a uniform field B, with the field perpendicular to the plane of the semicircle. The force on it equals that on:
(a) a straight wire of length πR
(b) a straight wire of length 2R
(c) a straight wire of length R
(d) zero
Show step-by-step solution
GivenSemicircular wire of radius R in a uniform field
AskedEquivalent straight wire
ConceptFor a bent wire in a uniform field, only the straight-line distance between the two ends matters.
FormulaF = I L_eff B, with L_eff = distance between the ends
SolutionThe two ends of a semicircle are separated by the diameter, 2R. So L_eff = 2R, not the arc length πR. The force equals that on a straight wire of length 2R.
Answer: a straight wire of length 2R
Q9Force on wire · concept
The net force on a closed current-carrying loop placed in a UNIFORM magnetic field is:
(a) BIL
(b) 2BIL
(c) zero
(d) depends on the shape
Show step-by-step solution
GivenClosed loop in a uniform field
AskedNet force on the loop
ConceptA closed loop begins and ends at the same point, so its effective length is zero.
FormulaF = I L_eff B, with L_eff = 0 for a closed loop
SolutionThe straight-line distance between the start and end of a closed loop is zero. So L_eff = 0 and the net force is zero. (A torque may still act — that is a different quantity.)
Answer: zero
Q10Force on wire · concept
In the formula F = I l × B, the magnetic field B refers to:
(a) the field produced by the wire itself
(b) an external magnetic field
(c) the sum of both
(d) the Earth's field only
Show step-by-step solution
GivenForce on a current-carrying conductor
AskedMeaning of B in the formula
ConceptA wire cannot exert a net force on itself, so the field must come from outside.
FormulaF = I l × B
SolutionThe wire does produce its own circular field, but that cannot push the wire itself. Just as you cannot lift yourself by pulling your own shoelaces. So B must be an external field.
Answer: an external magnetic field
Q11Force on wire · concept
In the expression F = I l × B, which quantity carries the direction information for the current?
(a) I
(b) l
(c) B
(d) F
Show step-by-step solution
GivenVector form of the force on a conductor
AskedThe vector that carries the current's direction
ConceptCurrent is a scalar quantity; the length vector is given the direction of current flow.
FormulaF = I l × B
SolutionCurrent I is not a vector — it is just a number of amperes. The vector l has magnitude equal to the length and points along the current. So l carries the direction information.
Answer: l
Q12Force on wire · numeric
A wire of length 1.5 m carries a current of 2 A and is suspended in mid-air by a horizontal magnetic field perpendicular to it. If the wire's mass is 200 g, the field is about:
(a) 0.33 T
(b) 0.65 T
(c) 1.3 T
(d) 2.6 T
Show step-by-step solution
GivenL = 1.5 m, I = 2 A, m = 200 g = 0.2 kg, g = 9.8 m/s²
AskedMagnetic field B
ConceptFor the wire to hang still, the upward magnetic force must exactly balance its weight.
FormulaB I L = m g ⇒ B = mg / IL
SolutionConvert mass first: 200 g = 0.2 kg. B = (0.2 × 9.8) / (2 × 1.5) = 1.96 / 3 = 0.65 T
Answer: 0.65 T
Q13Balance · numeric
A straight wire of mass 250 g and length 2 m is suspended horizontally in mid-air by a horizontal magnetic field of 0.7 T perpendicular to it. The current in the wire is:
(a) 1.25 A
(b) 1.75 A
(c) 2.5 A
(d) 3.5 A
Show step-by-step solution
Givenm = 250 g = 0.25 kg, L = 2 m, B = 0.7 T, g = 9.8 m/s²
AskedCurrent I
ConceptSame balance equation as before, rearranged for I this time.
FormulaB I L = m g ⇒ I = mg / BL
Solutionm = 250 g = 0.25 kg I = (0.25 × 9.8) / (0.7 × 2) = 2.45 / 1.4 = 1.75 A
Answer: 1.75 A
Q14Balance · numeric
A wire of mass per unit length 0.5 kg/m rests on a smooth inclined plane making 30° with the horizontal. A vertical magnetic field of 0.25 T prevents it from sliding down. The current required is about:
(a) 5.6 A
(b) 11.3 A
(c) 19.6 A
(d) 22.6 A
Show step-by-step solution
Givenm/L = 0.5 kg/m, α = 30°, B = 0.25 T (vertical), g = 9.8 m/s²
AskedCurrent I
ConceptOn a slope, the component of gravity down the incline is balanced by the component of the magnetic force along it. The tangent appears because the field is vertical.
FormulaI = (m/L) g tan α / B
Solutionmg sin α (down the slope) = BIL cos α (up the slope) I = (m/L) g tan α / B = (0.5 × 9.8 × tan 30°) / 0.25 = (4.9 × 0.577) / 0.25 = 2.83 / 0.25 = 11.3 A
Answer: 11.3 A
Q15Balance · concept
A wire is to be suspended in mid-air using a magnetic force. For this to be possible, the magnetic force must be:
(a) equal to mg and directed upwards
(b) equal to mg and directed downwards
(c) greater than mg
(d) any value, since the field supports it
Show step-by-step solution
GivenWire held motionless in mid-air
AskedCondition on the magnetic force
ConceptFor equilibrium the net force must be zero, so the upward push must exactly equal the downward pull.
FormulaB I L = m g
SolutionGravity pulls the wire down with force mg. For the wire to stay still, the magnetic force must be equal in size and opposite in direction. So it must equal mg and point upwards.
Answer: equal to mg and directed upwards
Q16Balance · ratio
A wire is suspended in mid-air by a magnetic field B carrying current I. If the current is doubled, the field required to keep it suspended becomes:
(a) double
(b) half
(c) four times
(d) unchanged
Show step-by-step solution
GivenSame wire (same m and L), current doubled
AskedNew field required
ConceptThe product BIL must still equal mg, so B and I are inversely related.
FormulaB I L = m g
Solutionmg is unchanged, so BIL must stay the same. If I doubles, B must halve to keep the product constant. B′ = B / 2
Answer: half
Q17Parallel wires · numeric
Two long parallel conductors 10 cm apart carry currents of 5 A and 10 A in the same direction. The force per unit length between them is:
(a) 1 × 10⁻⁴ N/m, attractive
(b) 1 × 10⁻⁴ N/m, repulsive
(c) 1 × 10⁻⁵ N/m, attractive
(d) 5 × 10⁻⁵ N/m, repulsive
Show step-by-step solution
GivenI₁ = 5 A, I₂ = 10 A, d = 10 cm = 0.1 m, same direction
AskedForce per unit length, and its nature
ConceptUse the two-wire formula, then decide attraction or repulsion from the current directions.
Formulaf = μ₀I₁I₂ / 2πd = (2 × 10⁻⁷) I₁I₂ / d
Solutionf = 2 × 10⁻⁷ × 5 × 10 / 0.1 = 2 × 10⁻⁷ × 50 / 0.1 = 10⁻⁵ / 0.1 = 1 × 10⁻⁴ N/m Currents are in the same direction ⇒ the wires ATTRACT.
Answer: 1 × 10⁻⁴ N/m, attractive
Q18Parallel wires · numeric
The same two wires (5 A and 10 A, 10 cm apart) now carry currents in opposite directions. The force per unit length is:
(a) 1 × 10⁻⁴ N/m, attractive
(b) 1 × 10⁻⁴ N/m, repulsive
(c) 2 × 10⁻⁴ N/m, repulsive
(d) zero
Show step-by-step solution
GivenSame currents and separation, opposite directions
AskedForce per unit length and its nature
ConceptThe size of the force does not change — only its direction does.
Formulaf = (2 × 10⁻⁷) I₁I₂ / d
SolutionThe magnitude is exactly as before: 1 × 10⁻⁴ N/m. Antiparallel currents REPEL. So the answer is 1 × 10⁻⁴ N/m, repulsive.
Answer: 1 × 10⁻⁴ N/m, repulsive
Q19Parallel wires · numeric
Two long parallel wires 6 cm apart carry currents of 2 A and 3 A. The force per unit length between them is:
(a) 1 × 10⁻⁵ N/m
(b) 2 × 10⁻⁵ N/m
(c) 4 × 10⁻⁵ N/m
(d) 6 × 10⁻⁵ N/m
Show step-by-step solution
GivenI₁ = 2 A, I₂ = 3 A, d = 6 cm = 0.06 m
AskedForce per unit length f
ConceptDirect substitution, using the shortcut μ₀/2π = 2 × 10⁻⁷.
Two long parallel wires carry currents in the same direction. They will:
(a) attract each other
(b) repel each other
(c) exert no force
(d) exert a torque only
Show step-by-step solution
GivenTwo parallel wires, currents in the same direction
AskedNature of the force
ConceptThis is the opposite of the rule for charges, so it has to be learnt as a fact.
FormulaParallel currents attract
SolutionCurrents flowing the same way produce fields that pull the wires together. So the wires ATTRACT. Note the contrast: two LIKE charges repel, but two LIKE currents attract.
Answer: attract each other
Q21Parallel wires · ratio
The force per unit length between two parallel wires is f. If the distance between them is doubled while the currents stay the same, the new force per unit length is:
(a) f / 2
(b) f / 4
(c) 2f
(d) 4f
Show step-by-step solution
Givend → 2d, same currents
AskedNew force per unit length
ConceptThe force falls off as 1/d — an inverse FIRST power, not inverse square.
Formulaf ∝ 1 / d
Solutionf′ / f = d / 2d = 1/2 f′ = f / 2
Answer: f / 2
Q22Parallel wires · ratio
If both currents in a pair of parallel wires are doubled while the separation stays the same, the force per unit length becomes:
(a) double
(b) four times
(c) half
(d) unchanged
Show step-by-step solution
GivenI₁ → 2I₁ and I₂ → 2I₂, same d
AskedNew force per unit length
ConceptBoth currents appear as factors, so doubling each multiplies the force by 2 × 2.
ConceptThe ampere is a foundation stone of the SI system, alongside the metre, kilogram and second.
FormulaDefinition via the force between two wires
SolutionThe ampere is one of the seven SI BASE units. The coulomb is defined FROM the ampere (1 C = 1 A × 1 s), not the other way round.
Answer: one of the seven SI base units
Q25Coulomb definition
One coulomb is the charge that flows through a cross-section when:
(a) 1 A flows for 1 s
(b) 1 A flows for 1 minute
(c) 1 V is applied for 1 s
(d) 1 N of force acts for 1 s
Show step-by-step solution
GivenDefinition of the coulomb from the ampere
AskedCorrect statement
ConceptOnce the ampere is defined, charge follows as current multiplied by time.
FormulaQ = I t
SolutionQ = I t With I = 1 A and t = 1 s, Q = 1 C.
Answer: 1 A flows for 1 s
Q26Parallel wires · Newton's third law
For two long parallel current-carrying wires, the force exerted by wire A on wire B compared with that by B on A is:
(a) greater if A carries more current
(b) smaller if A carries more current
(c) equal in magnitude and opposite in direction
(d) zero
Show step-by-step solution
GivenTwo parallel wires carrying steady currents
AskedComparison of the mutual forces
ConceptThe formula contains the product I₁I₂, which is symmetric — so both wires feel the same size of force even if their currents differ.
FormulaF_ab = − F_ba
SolutionThe expression μ₀I₁I₂L/2πd is unchanged if the two currents are swapped. So both wires feel forces of equal magnitude, in opposite directions. This is consistent with Newton's third law.
Answer: equal in magnitude and opposite in direction
Q27Levitation
Two infinite parallel wires A and B carry currents I and 2I in the same direction. Wire A has mass per unit length λ and lies on the floor; B is fixed at a height h directly above it. For A to just remain on the floor, h must satisfy:
(a) h ≥ μ₀I² / πλg
(b) h ≥ μ₀I² / 2πλg
(c) h ≤ μ₀I² / πλg
(d) h ≥ 2μ₀I² / πλg
Show step-by-step solution
GivenCurrents I and 2I in the same direction, mass per unit length λ, separation h
AskedCondition on h
ConceptSame-direction currents attract, so B pulls A upwards. For A to stay down, that upward pull must not exceed its weight per unit length.
Formulaμ₀I₁I₂ / 2πh ≤ λ g
SolutionUpward pull per unit length = μ₀(I)(2I) / 2πh = μ₀I² / πh Weight per unit length = λg For A to stay on the floor: μ₀I²/πh ≤ λg Rearranging: h ≥ μ₀I² / πλg
Answer: h ≥ μ₀I² / πλg
Q28Levitation · concept
A wire is to float in mid-air directly above another fixed current-carrying wire. For this to happen, the currents must be:
(a) in the same direction
(b) in opposite directions
(c) equal in magnitude
(d) perpendicular to each other
Show step-by-step solution
GivenOne wire floating above another
AskedRequired current directions
ConceptThe floating wire needs an upward force, and only attraction can pull it up towards the wire above it.
FormulaParallel currents attract
SolutionThe upper wire must PULL the lower one up, or the lower must be pushed up by the one above. If the wire floats BELOW the fixed one, it must be attracted upwards ⇒ same direction. Only parallel (same-direction) currents attract.
Answer: in the same direction
Q29Three wires
Three long parallel wires lie in a plane, equally spaced, and all carry equal currents in the same direction. The net force per unit length on the MIDDLE wire is:
(a) zero
(b) μ₀I²/2πd
(c) μ₀I²/πd
(d) 2μ₀I²/πd
Show step-by-step solution
GivenThree equally spaced parallel wires, equal currents, same direction
AskedNet force on the middle wire
ConceptEach outer wire attracts the middle one, but towards opposite sides. Equal pulls in opposite directions cancel.
Formulaf = μ₀I₁I₂ / 2πd, added as vectors
SolutionThe left wire attracts the middle wire to the left. The right wire attracts it to the right, with equal magnitude. The two pulls are equal and opposite, so they cancel: net force = zero.
Answer: zero
Q30Two wires · perpendicular
Two long straight wires are placed perpendicular to each other and carry currents. The net force between them is:
ConceptThe simple parallel-wire formula applies only to parallel segments. For perpendicular wires the forces on opposite halves cancel.
FormulaF = I l × B, integrated over the wire
SolutionFor perpendicular wires, one half of each wire is pushed one way and the other half the opposite way. These contributions cancel in pairs. The net force is zero, although a torque can exist.
Answer: zero
Q31Loop near a wire
A square loop carrying current i lies in the same plane as a long straight wire carrying current I, with one arm parallel and closest to the wire. If the near arm's current is in the same direction as I, the loop experiences:
(a) no net force
(b) a net force towards the wire
(c) a net force away from the wire
(d) a torque only
Show step-by-step solution
GivenSquare loop coplanar with a long straight wire, near arm parallel to it
AskedNet force on the loop
ConceptThe field of the wire is NON-uniform, so the zero-net-force rule does not apply. Compare the near and far arms.
FormulaF = μ₀Ii L / 2πd on each parallel arm
SolutionThe two arms perpendicular to the wire feel equal and opposite forces that cancel. The near arm (same direction as I) is attracted; the far arm (opposite direction) is repelled. Force falls off as 1/d, so the nearer arm feels the greater force. Net result: the loop is pulled TOWARDS the wire.
Answer: a net force towards the wire
Q32Loop near a wire · numeric
A square loop of side 10 cm carrying 2 A lies with its near arm 10 cm from a long wire carrying 10 A, both in the same plane and the near arm parallel to the wire. The magnitude of the net force on the loop is:
(a) 1 × 10⁻⁵ N
(b) 2 × 10⁻⁵ N
(c) 4 × 10⁻⁵ N
(d) zero
Show step-by-step solution
GivenI = 10 A, i = 2 A, a = 0.1 m, x = 0.1 m
AskedNet force on the loop
ConceptOnly the two arms parallel to the wire contribute. Compute each and subtract, since one is attracted and the other repelled.
Two long parallel wires carry currents I and 4I in the same direction and are 30 cm apart. A third wire placed parallel to them experiences no net force when it is placed at a distance from the first wire of:
(a) 6 cm
(b) 10 cm
(c) 15 cm
(d) 24 cm
Show step-by-step solution
GivenI₁ = I, I₂ = 4I, separation = 30 cm, same direction
AskedPosition of the third wire for zero net force
ConceptThe third wire feels a pull towards each of the others. For these to cancel it must sit between them, where the two forces are equal.
Formulaμ₀I₁I₃ / 2πx = μ₀I₂I₃ / 2π(d − x)
SolutionCancel the common factors: I₁/x = I₂/(30 − x) 1/x = 4/(30 − x) 30 − x = 4x ⇒ 5x = 30 ⇒ x = 6 cm It lies nearer the weaker current, as expected.
Answer: 6 cm
Q34Parallel wires · derivation
The force between two parallel current-carrying wires is obtained by combining:
(a) Ampere's law and Ohm's law
(b) the field of a straight wire and the force on a conductor
(c) Biot–Savart law and Coulomb's law
(d) the torque formula and Newton's third law
Show step-by-step solution
GivenDerivation of f = μ₀I₁I₂/2πd
AskedWhich two results are combined
ConceptWire A's field is found first, then the force it exerts on wire B is computed.
FormulaB = μ₀I₁/2πd, then F = B I₂ L
SolutionStep 1: the field of wire A at wire B is B = μ₀I₁/2πd. Step 2: the force on wire B in that field is F = I₂LB. Combining: F = μ₀I₁I₂L/2πd. So it uses the straight-wire field plus the force-on-a-conductor formula.
Answer: the field of a straight wire and the force on a conductor
Q35Parallel wires · comparison
Which statement correctly compares currents with charges?
(a) Like charges attract; like currents attract
(b) Like charges repel; like currents attract
(c) Like charges repel; like currents repel
(d) Like charges attract; like currents repel
Show step-by-step solution
GivenComparison of electrostatic and magnetic behaviour
AskedThe correct comparison
ConceptThis reversal is one of the most commonly examined contrasts in the chapter.
FormulaCoulomb's law vs f = μ₀I₁I₂/2πd
SolutionTwo like (same sign) charges REPEL. Two like (same direction) currents ATTRACT. So the correct pairing is: like charges repel, like currents attract.
Answer: Like charges repel; like currents attract
Q36Parallel wires · numeric
Two parallel wires 1 m apart each carry 10 A in the same direction. The force on a 2 m length of one wire is:
(a) 2 × 10⁻⁶ N
(b) 4 × 10⁻⁵ N
(c) 2 × 10⁻⁵ N
(d) 4 × 10⁻⁶ N
Show step-by-step solution
GivenI₁ = I₂ = 10 A, d = 1 m, L = 2 m, same direction
AskedTotal force on a 2 m length
ConceptCompute the force per metre first, then multiply by the length asked for.
FormulaF = μ₀I₁I₂L / 2πd
Solutionf = 2 × 10⁻⁷ × 10 × 10 / 1 = 2 × 10⁻⁵ N/m F = f × L = 2 × 10⁻⁵ × 2 = 4 × 10⁻⁵ N, attractive
Answer: 4 × 10⁻⁵ N
Q37Parallel wires · concept
The force per unit length between two parallel wires varies with their separation d as:
(a) 1 / d
(b) 1 / d²
(c) d
(d) d²
Show step-by-step solution
GivenTwo long parallel current-carrying wires
AskedDependence on separation
ConceptRead it off the formula. This is a 1/d law, not an inverse-square law.
Formulaf = μ₀I₁I₂ / 2πd
Solutiond appears once in the denominator. So f ∝ 1/d. (Contrast with Coulomb's law between charges, which is 1/d².)
Answer: 1 / d
Q38Comparison
A wire lies in a magnetic field. Which of the following does NOT affect the force on it?
(a) the current in the wire
(b) the length of the wire
(c) the resistance of the wire
(d) the angle between the wire and the field
Show step-by-step solution
GivenForce on a current-carrying conductor
AskedThe quantity that does not appear
ConceptCheck the formula and see which of the four is absent from it.
FormulaF = B I L sin θ
SolutionThe formula contains B, I, L and sin θ. Resistance does not appear anywhere in it. (Resistance affects how much current flows, but for a GIVEN current it does not affect the force.)
Answer: the resistance of the wire
Q39Comparison
Which of the following has units of newtons per metre?
(a) μ₀I / 2πr
(b) B I L sin θ
(c) μ₀I₁I₂ / 2πd
(d) μ₀ n I
Show step-by-step solution
GivenFour expressions from the chapter
AskedThe one with units N/m
ConceptChecking the unit is the fastest way to identify which formula an option belongs to.
Formulaf = μ₀I₁I₂ / 2πd
Solutionμ₀I/2πr is a magnetic field ⇒ tesla. BIL sin θ is a force ⇒ newton. μ₀nI is a magnetic field ⇒ tesla. μ₀I₁I₂/2πd is force per unit length ⇒ newtons per metre.
Answer: μ₀I₁I₂ / 2πd
Q40Concept · non-uniform field
The statement 'the net force on a closed current loop is zero' is true only if the magnetic field is:
(a) strong
(b) weak
(c) uniform
(d) perpendicular to the loop
Show step-by-step solution
GivenClosed loop in a magnetic field
AskedThe necessary condition
ConceptThe cancellation of forces on opposite arms relies on both sides sitting in the same field strength.
FormulaF_net = ∑ I dl × B
SolutionIn a uniform field, opposite arms feel equal and opposite forces and cancel. In a non-uniform field the arms sit in different field strengths, so the cancellation fails. Hence the statement requires a UNIFORM field.
Answer: uniform
Q41Concept · effective length
A wire of arbitrary shape carries current I between two points 40 cm apart in a uniform field of 0.5 T perpendicular to the plane. If I = 3 A, the force on the wire is:
(a) 0.3 N
(b) 0.6 N
(c) 1.2 N
(d) depends on the wire's shape
Show step-by-step solution
GivenEnd-to-end distance = 40 cm = 0.4 m, I = 3 A, B = 0.5 T, θ = 90°
AskedForce on the wire
ConceptFor a bent wire in a uniform field, only the straight-line distance between the two ends matters.
FormulaF = I L_eff B sin θ
SolutionL_eff = 0.4 m (the straight line joining the ends). F = 3 × 0.4 × 0.5 × 1 = 0.6 N The actual shape of the wire is irrelevant.
Answer: 0.6 N
Q42Assertion–Reason
Assertion (A): Two parallel wires carrying currents in the same direction attract each other. Reason (R): Each wire lies in the magnetic field produced by the other.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show step-by-step solution
GivenTwo parallel wires with currents in the same direction
AskedTruth of A and R, and whether R explains A
ConceptCheck both statements, then ask whether the reason actually produces the attraction.
FormulaB = μ₀I₁/2πd, then F = I₂LB
SolutionA is TRUE: parallel currents attract. R is TRUE: each wire does sit in the field created by the other. And that is precisely the mechanism — wire A's field acts on wire B's current, producing the force. So R correctly explains A.
Answer: Both A and R are true, and R is the correct explanation of A
Q43Assertion–Reason
Assertion (A): The net force on a current loop placed in a non-uniform magnetic field is zero. Reason (R): Forces on opposite arms of a loop are always equal and opposite.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) Both A and R are false
Show step-by-step solution
GivenCurrent loop in a non-uniform field
AskedTruth of A and R
ConceptBoth statements quietly assume uniformity, which is exactly what is missing here.
FormulaF_net = ∑ I dl × B
SolutionA is FALSE: in a NON-uniform field the net force is generally not zero. R is FALSE: the forces on opposite arms are equal and opposite only in a UNIFORM field. So both statements are false.
Answer: Both A and R are false
Q44Assertion–Reason
Assertion (A): A current-carrying wire placed along a magnetic field experiences no force. Reason (R): The force on a conductor is proportional to sin θ, where θ is the angle between the wire and the field.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show step-by-step solution
GivenWire lying along the field direction
AskedTruth of A and R
ConceptBoth follow directly from the force formula.
FormulaF = B I L sin θ
SolutionA is TRUE: with θ = 0, sin θ = 0 and the force vanishes. R is TRUE: the force does contain a sin θ factor. And that factor is exactly why the force is zero, so R explains A.
Answer: Both A and R are true, and R is the correct explanation of A
Q45Match the following
Match each situation with the correct result:
(a) Wire perpendicular to B (b) Wire parallel to B (c) Closed loop in a uniform field (d) Two antiparallel currents
(i) Zero force (ii) Maximum force BIL (iii) Zero net force (iv) Repulsion
(a) a–ii, b–i, c–iii, d–iv
(b) a–i, b–ii, c–iii, d–iv
(c) a–ii, b–iii, c–i, d–iv
(d) a–ii, b–i, c–iv, d–iii
Show step-by-step solution
GivenFour standard situations
AskedCorrect matching
ConceptWork through each with the appropriate formula.
FormulaF = BIL sin θ; f = μ₀I₁I₂/2πd
Solution(a) Perpendicular ⇒ sin 90° = 1 ⇒ maximum force BIL ⇒ (ii) (b) Parallel ⇒ sin 0° = 0 ⇒ zero force ⇒ (i) (c) Closed loop in a uniform field ⇒ zero NET force ⇒ (iii) (d) Antiparallel currents ⇒ repulsion ⇒ (iv)
Answer: a–ii, b–i, c–iii, d–iv
Q46Match the following
Match each expression with what it represents:
(a) μ₀I / 2πr (b) BIL sin θ (c) μ₀I₁I₂ / 2πd (d) 2 × 10⁻⁷ N/m
(i) Force on a conductor (ii) Field of a straight wire (iii) Basis of the ampere's definition (iv) Force per unit length between two wires
(a) a–ii, b–i, c–iv, d–iii
(b) a–i, b–ii, c–iv, d–iii
(c) a–ii, b–iv, c–i, d–iii
(d) a–iii, b–i, c–iv, d–ii
Show step-by-step solution
GivenFour expressions from sections 4.2.3, 4.6 and 4.8
AskedCorrect matching
ConceptIdentifying each expression by its unit is the fastest route.
FormulaStandard formulas
Solution(a) μ₀I/2πr ⇒ field of a long straight wire ⇒ (ii) (b) BIL sin θ ⇒ force on a conductor ⇒ (i) (c) μ₀I₁I₂/2πd ⇒ force per unit length between two wires ⇒ (iv) (d) 2 × 10⁻⁷ N/m ⇒ the value used to define the ampere ⇒ (iii)
Answer: a–ii, b–i, c–iv, d–iii
Q47Match the following
Match each balance situation with the correct expression:
(a) Wire suspended in mid-air (b) Rod on an incline α (c) Wire floating above another (d) Closed loop in a uniform field
(i) μ₀I₁I₂/2πh = λg (ii) BIL = mg (iii) F = 0 (iv) I = (m/L)g tan α / B
(a) a–ii, b–iv, c–i, d–iii
(b) a–ii, b–i, c–iv, d–iii
(c) a–iv, b–ii, c–i, d–iii
(d) a–ii, b–iv, c–iii, d–i
Show step-by-step solution
GivenFour equilibrium situations
AskedCorrect matching
ConceptEvery one of these comes from the same idea: write down what pushes up and what pulls down, then set them equal.
FormulaBalance of forces
Solution(a) Wire in mid-air ⇒ BIL = mg ⇒ (ii) (b) Rod on an incline ⇒ I = (m/L)g tan α / B ⇒ (iv) (c) Wire floating above another ⇒ μ₀I₁I₂/2πh = λg ⇒ (i) (d) Closed loop in a uniform field ⇒ net force zero ⇒ (iii)
Answer: a–ii, b–iv, c–i, d–iii
Q48Recognition
A question states: 'Two long parallel conductors 20 cm apart carry 4 A and 6 A. Find the force on each metre of length.' Which formula applies?
(a) B = μ₀I / 2πr
(b) F = BIL sin θ
(c) f = μ₀I₁I₂ / 2πd
(d) τ = mB sin θ
Show step-by-step solution
GivenTwo parallel conductors, currents 4 A and 6 A, 20 cm apart, force per metre
AskedThe correct formula
ConceptTwo currents and a separation, with the answer wanted per metre — that is the two-wire formula.
Formulaf = (2 × 10⁻⁷) I₁I₂ / d
SolutionThe question gives TWO currents and a separation ⇒ two-wire formula. It asks for force per unit length ⇒ answer in N/m. f = 2 × 10⁻⁷ × 4 × 6 / 0.2 = 2 × 10⁻⁷ × 120 = 2.4 × 10⁻⁵ N/m
Answer: f = μ₀I₁I₂ / 2πd
Q49Concept · summary
Which of the following statements is INCORRECT?
(a) A wire parallel to a magnetic field experiences no force
(b) Two antiparallel currents repel each other
(c) The net force on a closed loop in a uniform field is zero
(d) The force between two parallel wires varies as the inverse square of their separation
Show step-by-step solution
GivenStatements about forces on conductors
AskedThe incorrect statement
ConceptCheck each against the relevant formula. The distance dependence is the one that is commonly misremembered.
FormulaF = BIL sin θ; f = μ₀I₁I₂/2πd
Solution'Parallel wire feels no force' — correct, sin 0° = 0. 'Antiparallel currents repel' — correct. 'Net force on a closed loop in a uniform field is zero' — correct. 'Force varies as inverse square of separation' — INCORRECT. It varies as 1/d, not 1/d².
Answer: The force between two parallel wires varies as the inverse square of their separation
Q50Balance · numeric
A horizontal wire of length 0.5 m and mass 20 g carries a current I in a horizontal magnetic field of 0.4 T perpendicular to it. The current needed to just support the wire against gravity is:
(a) 0.98 A
(b) 1.96 A
(c) 2.45 A
(d) 4.9 A
Show step-by-step solution
GivenL = 0.5 m, m = 20 g = 0.02 kg, B = 0.4 T, g = 9.8 m/s²
AskedCurrent I
ConceptThe magnetic force must exactly balance the weight of the wire.
FormulaB I L = m g ⇒ I = mg / BL
SolutionConvert the mass: 20 g = 0.02 kg. Weight = mg = 0.02 × 9.8 = 0.196 N I = 0.196 / (0.4 × 0.5) = 0.196 / 0.2 I = 0.98 A
Answer: 0.98 A
Before the next ILTS — how to use this pack
Read Part 1 once, slowly. The two halves of this topic are the same physics used twice — first a wire in someone else's field, then two wires in each other's.
Copy the formula sheet by hand, then cover it and rewrite it from memory.
Q1–Q11 (force on a wire) in one sitting. Mostly one-line substitutions plus direction work.
Q12–Q16 (the balance family) next, and treat these as the priority block — three separate NEET years reduce to this one method.
Q17–Q36 (parallel wires) the following day, including levitation, three-wire and loop-near-a-wire cases.
Q37–Q50 are mixed, assertion–reason and matching. Save them for final revision.
The four mistakes that cost the most marks in this topic
Treating the two-wire force as inverse-square. It goes as 1/d, not 1/d².
Getting attraction and repulsion the wrong way round. Like charges repel, but like currents attract.
Forgetting to convert grams to kilograms in the balance questions. 250 g is 0.25 kg.
Applying "net force on a closed loop is zero" to a loop sitting near a straight wire. That field is not uniform, so the rule does not hold.
The one-minute self-test. If she can answer these four without hesitating, this topic is secure:
A wire hangs in mid-air in a magnetic field. What equation do you write down first?
Two wires carry current the same way. Do they attract or repel — and how is that different from charges?
Why is the net force on a closed loop zero in a uniform field but not near a straight wire?
Where does the number 2 × 10⁻⁷ in the definition of the ampere come from?