1. Motion in each magnetic region

A charged particle moving with velocity perpendicular to a uniform magnetic field moves in a circle of radius

$r = \frac{mv}{qB}$

So, in the two regions:

• In region 1: $r_1 = \frac{mv}{qB_1}$
• In region 2: $r_2 = \frac{mv}{qB_2}$

Given $B_2 > B_1$, we have

$r_2 < r_1$

2. Path of the particle

The particle starts at the interface and enters region 2 with velocity along the interface-normal direction. Since magnetic force is always perpendicular to velocity, the particle follows a circular arc in region 2 and returns to the interface.

Because it starts from the interface and comes back to the interface, the motion in each region is a semicircle.

Hence:

• In region 2, it completes a semicircle of radius $r_2$ and returns to the interface.
• Then it enters region 1, again moving along a semicircle of radius $r_1$, and again reaches the interface.

3. Displacement along the interface

For a semicircular path starting and ending on the interface, the shift along the interface equals the diameter of the circular path.

Therefore:

• Shift along interface in region 2 = $2r_2$
• Shift along interface in region 1 = $2r_1$

The magnetic field directions on the two sides cause the particle to bend on opposite sides of the normal, so these two shifts are in opposite directions along the interface.

Thus net displacement along the interface is

$2r_1 - 2r_2 = 2(r_1-r_2)$

Substitute $r_1$ and $r_2$:

$\Delta x = 2\left(\frac{mv}{qB_1}-\frac{mv}{qB_2}\right)$

Factor out $\frac{mv}{qB_1}$:

$\Delta x = \frac{mv}{qB_1}\left(1-\frac{B_1}{B_2}\right) \times 2$

4. Match with options

This matches Option A.

$\boxed{\Delta x = \frac{mv}{qB_1}\left(1-\frac{B_1}{B_2}\right)\times 2}$