JEE Main 2025 Apr 7 Shift 1, Physics Q4: Motion of a Charged Particle in a Magnetic Field

JEE Main2025Apr 7, Shift 1Physics

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 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 $B_{2} > B_{1}$ )

A
$\frac{m v}{q B _{1}} (1 - \frac{B _{2}}{B _{1}}) \times 2$
B
$\frac{m v}{q B _{1}} (1 - \frac{B _{1}}{B _{2}})$
C
$\frac{m v}{q B _{1}} (1 - \frac{B _{2}}{B _{1}})$
D
$\frac{m v}{q B _{1}} (1 - \frac{B _{1}}{B _{2}}) \times 2$

Solution

Since the velocity is perpendicular to the magnetic field, the particle moves in a circular arc in each region. The radius of the circle in a field $B$ is $R = \frac{m v}{q B}$ .

In region 2 (field $B_{2}$ ): the particle traces a semicircle and returns to the interface. The chord traversed along the interface equals the diameter:

$d_{2} = 2 R_{2} = \frac{2 m v}{q B _{2}}$

In region 1 (field $B_{1}$ ): the particle continues on the same speed (magnetic force does no work), traces another semicircle, and returns to the interface. Its chord along the interface:

$d_{1} = 2 R_{1} = \frac{2 m v}{q B _{1}}$

Net displacement along the interface. The two arcs curl in opposite senses about the interface, so the two chord-displacements are directed opposite to each other. The net magnitude is

$∣ d_{1} - d_{2} ∣ = \frac{2 m v}{q B _{1}} - \frac{2 m v}{q B _{2}} = \frac{2 m v}{q B _{1}} (1 - \frac{B _{1}}{B _{2}})$

$∴$ Option (4).

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JEE Main 2025 Apr 7 Shift 1, Physics Q4: Motion of a Charged Particle in a Magnetic Field

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