Four Doubts,
Solved Fully

1Given

• n turns, wound tightly as a flat spiral
• Inner radius r₁, outer radius r₂ (with r₂ > r₁)
• Current I in every turn
• Implicit: the turns are spread evenly across the radial width (r₂ − r₁)
• Implicit: each turn is close enough to a full circle to use the loop formula

2Asked

The magnetic field B at the centre of the spiral.

3Concept

One circular loop of radius r gives a field at its centre of B = μ0I / 2r. But here every turn sits at a different radius, so there is no single r to plug in.

So do what Biot–Savart always asks: chop the spiral into thin rings, work out what one ring gives, and add them all up. Every turn carries current the same way round, so all their fields point the same way and simply add as numbers — no vectors needed.

4Method & baby steps

1. How many turns lie in a thin ring between r and r + dr?
   n turns are spread over a width of (r₂ − r₁) so turns per unit width = n / (r₂ − r₁) in a slice of width dr: dN = n dr / (r₂ − r₁)
2. What field does that thin ring make?
   one loop of radius r gives μ₀I / 2r dN loops of radius r give dB = μ₀ I dN / 2r substitute dN: dB = μ₀ n I dr / 2r(r₂ − r₁)
3. Add up every ring from r₁ out to r₂
   pull the constants outside: B = μ₀nI / 2(r₂ − r₁) × ∫ dr/r from r₁ to r₂
4. Do the integral — this is the only one that gives a log
   ∫ dr/r = ln r evaluated: ln r₂ − ln r₁ = ln(r₂/r₁)
5. Put it together
   B = μ₀nI / 2(r₂ − r₁) · ln(r₂/r₁)

5Easy tricks & shortcuts

6Solution

7Diagram

1Given

• Electron: charge e = 1.6 × 10⁻¹⁹ C, mass m = 9.1 × 10⁻³¹ kg
• Initial kinetic energy K = 100 eV
• Field B = 1.5 × 10⁻³ T, along k̂ (out of the page)
• Field exists only between x = 0 and x = 2 cm
• Screen is 8 cm from S → so the field-free drift is 8 − 2 = 6 cm
• Implicit: P is where the electron would have landed undeflected

2Asked

The distance d = PQ measured on the screen.

3Concept

The journey has two quite different halves, and the whole difficulty of the question is remembering that.

• Inside the field (2 cm): a circular arc of radius r = mv/eB. The electron gains both a sideways shift and a slope.
• After the field (6 cm): no force at all, so a perfectly straight line at whatever slope it left with.

The total deflection is the sum of the two. Forget the first part and you land on one of the wrong options that has been placed there deliberately.

4Method & baby steps

1. Turn the energy into a speed
   K = 100 eV = 100 × 1.6×10⁻¹⁹ = 1.6 × 10⁻¹⁷ J v = √(2K/m) = √(2 × 1.6×10⁻¹⁷ ÷ 9.1×10⁻³¹) = √(3.52 × 10¹³) = 5.93 × 10⁶ m/s
2. Find the radius of the arc
   r = mv / eB = (9.1×10⁻³¹ × 5.93×10⁶) ÷ (1.6×10⁻¹⁹ × 1.5×10⁻³) = 5.40×10⁻²⁴ ÷ 2.40×10⁻²² = 0.0225 m = 2.25 cm
3. Find the angle it has turned through on leaving the field
   horizontal distance covered in the field = 2 cm sin θ = 2 / 2.25 = 0.889 θ = 62.8° cos θ = 0.458 tan θ = 1.94
4. Deflection gained INSIDE the field
   y₁ = r − r cos θ = r(1 − cos θ) = 2.25 × (1 − 0.458) = 2.25 × 0.542 = 1.22 cm
5. Deflection gained while drifting the remaining 6 cm
   y₂ = (8 − 2) × tan θ = 6 × 1.94 = 11.65 cm
6. Add the two
   d = y₁ + y₂ = 1.22 + 11.65 = 12.87 cm

5Easy tricks & shortcuts

6Solution

7Diagram

1Given

• Region 1 (a < x < 2a): B = +B₀ ĵ
• Region 2 (2a < x < 3a): B = −B₀ ĵ
• Charge is positive
• Enters at x = a with v = v₀ î (purely along +x)
• The graphs are drawn in the x–z plane

2Asked

Which of the four sketched trajectories is correct.

3Concept

Two ideas settle this without any calculation.

• Direction. F = q(v × B). With v along x and B along y, the force is along z. Get its sign and half the options die.
• Reversal. Flipping B flips the direction of turning, not the sign of the velocity the particle has already built up. So the curve changes its bend, but z carries on increasing.

Also worth holding on to: the magnetic force does no work, so the speed never changes — only the direction.

4Method & baby steps

1. Get the initial force direction
   F = q (v × B) = q (v₀ î × B₀ ĵ) = q v₀ B₀ (î × ĵ) and î × ĵ = k̂, so F is along +z — the particle curves UPWARD
2. Eliminate immediately
   options (b) and (d) show the path going DOWN → both are out
3. What happens across Region 1 (a to 2a)
   velocity rotates steadily from +x toward +z vz grows from 0 upward so the curve gets steeper — concave up
4. What happens across Region 2 (2a to 3a)
   B has flipped sign, so the turning reverses velocity now rotates back from +z toward +x vz shrinks toward zero — but stays POSITIVE so the curve flattens — concave down
5. Does z ever come back down?
   z would only fall if vz went NEGATIVE reversing B reverses the RATE of turning, not vz itself vz only decays back toward zero so z rises the whole way — option (c) is out
6. The surviving shape
   rises, steepens, inflects at 2a, then flattens an S-curve that keeps climbing → option (a)

5Easy tricks & shortcuts

6Solution

7Diagram

1Given

• Mass m, charge q, speed v
• Enters Region II normal to the boundary — this word is the key to the whole question
• Region II: uniform B, perpendicular to the page, width ℓ
• Regions I and III are field free
• Implicit: the speed never changes, since a magnetic force does no work

2Asked

Which of the four statements (a)–(d) are correct. More than one may be.

3Concept

Inside Region II the particle travels on a circle of radius r = mv/qB. Everything follows from one geometric fact:

Because it enters normal to the boundary, the centre of that circle lies on the boundary itself. So the deepest the particle can ever penetrate is exactly one radius. Compare r with ℓ and every part of the question answers itself.

4Method & baby steps

1. Fix the geometry
   r = mv / qB entering normal ⇒ centre of the circle sits ON the entry boundary ⇒ maximum penetration depth = exactly r
2. Condition to reach Region III → tests (a) and (b)
   needs r > ℓ mv / qB > ℓ v > qℓB / m so (a) is CORRECT and (b) is WRONG
3. Path length inside Region II → tests (c)
   if r < ℓ : it does a HALF circle and returns path length = πr, which grows with r largest as r → ℓ : πℓ ≈ 3.14 ℓ if r > ℓ : it turns through θ with sin θ = ℓ/r path length = rθ as r → ℓ from above: θ → 90°, path → 1.57 ℓ so the path length is longest exactly at r = ℓ v = qℓB / m → (c) is CORRECT
4. Time spent when it comes back → tests (d)
   coming back means a complete half circle t = T/2 = (1/2)(2πm / qB) = πm / qB no v anywhere in it → same for every speed that returns → (d) is CORRECT

5Easy tricks & shortcuts

6Solution

7Diagram