JEE MAIN - Physics (2025 - 7th April Morning Shift - No. 11)
Uniform magnetic fields of different strengths $\left(B_1\right.$ and $\left.B_2\right)$, 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?
_7th_April_Morning_Shift_en_11_1.png)
(Consider the velocity of the particle to be normal to the magnetic field and $\mathrm{B}_2>\mathrm{B}_1$ )
Explanation
As $\overrightarrow{\mathrm{v}}$ is $\perp$ to $\overrightarrow{\mathrm{B}}$, so charge particle will move in circular path, whose radius is given by
$$\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}$$
_7th_April_Morning_Shift_en_11_2.png)
Starting point $\rightarrow \mathrm{A}$
Ending point $\rightarrow \mathrm{C}$
$\therefore$ Net displacement $=\mathrm{AC}$
$$\begin{aligned} & \mathrm{AC}=\mathrm{CD}-\mathrm{AD} \\ & \mathrm{AC}=\frac{2 \mathrm{mv}}{\mathrm{qB}_1}-\frac{2 \mathrm{mv}}{\mathrm{qB}_2} \\ & \mathrm{AC}=\frac{2 \mathrm{mv}}{\mathrm{qB}_1}\left[1-\frac{\mathrm{B}_1}{\mathrm{~B}_2}\right] \end{aligned}$$
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