Uniform magnetic fields of different strengths (B₁ and B₂), both normal to the plane of the paper exist as shown in the figure. A charged particle of mass m and charge q, at the interface at an instant, moves into the region 2 with velocity v and returns to the interface. It continues to move into region 1 and finally reaches the interface. What is the displacement of the particle during this movement along the interface?
(Consider the velocity of the particle to be normal to the magnetic field and B₂>B₁ )

Solution

Physics sir · 4 steps

Two half-circles that curl the same way: one diameter pushes you back, the other pushes you forward.

1. Radius in each region. The speed stays v because a magnetic force does no work. So the radius is r = mv/qB, giving r₂ = mv/qB₂ in region 2 and r₁ = mv/qB₁ in region 1. Since B₂ > B₁, the circle in region 2 is smaller.
2. Half-circle each time. Take the particle as entering normal to the interface. In region 2 it traces a semicircle and comes back to the interface one diameter, 2r₂, along the line. It then enters region 1 and traces a semicircle of diameter 2r₁.
3. Opposite shifts. Both fields point into the paper, so the particle turns the same way (say clockwise) in both regions. Going down and turning clockwise shifts it one way along the interface. Going up and turning clockwise shifts it the other way. The two diameters therefore subtract.

   d = 2r₁ - 2r₂ = 2mv/qB₁ - 2mv/qB₂ = mv/qB₁(1 - B₁/B₂) × 2

4. So. The net displacement along the interface is mv/qB₁(1 - B₁/B₂) × 2, which is option (a). It is positive because B₂ > B₁ makes r₁ > r₂.

The idea to keep. Same field direction on both sides means the same sense of rotation, so the two diameters along the interface point opposite ways. The net drift is their difference.

Answer from the NTA final answer key. Solution by PrepWiser.