Question

Uniform magnetic fields of different strengths $ B_1 $ and $ B_2 $, 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 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  $ B_2 > B_1 $.

• A. \( \frac{mv}{qB_1} \left( 1 - \frac{B_2}{B_1} \right) \times 2 \)
• B. \( \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \)
• C. \( \frac{mv}{qB_1} \left( 1 - \frac{B_2}{B_1} \right) \)
• D. \( \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \times 2 \)

Hint:

For problems involving charged particles in magnetic fields, remember that the velocity of the particle and the magnetic field strength determine the radius of the circular motion. Use this to relate the displacement.

Answer 1

The Correct Option is D

A charged particle traverses between two regions with differing magnetic field strengths, \( B_1 \) and \( B_2 \), returning to its origin. The particle's motion is governed by the Lorentz force, resulting in circular paths within each region. The radius of curvature is dependent on the particle's velocity \( v \), charge \( q \), and the local magnetic field strength. The observed displacement along the interface is directly proportional to the difference between the magnetic fields. Specifically, the displacement is given by the formula: \[ \text{Displacement} = \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \times 2. \] Consequently, the derived displacement is: \[ \frac{mv}{qB_1} \left( 1 - \frac{B_1}{B_2} \right) \times 2. \]