JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2025
- Easy
- MCQ
Uniform magnetic fields of different strengths \(\left(B_1\right.\) 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 \(B_2 \gt B_1\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{mv}}{\mathrm{qB}_1}\left(1-\frac{\mathrm{B}_1}{\mathrm{~B}_2}\right) \times 2\)
Step-by-step Solution
Detailed explanation
As \(\overrightarrow{\mathrm{v}}\) is \(\perp\) to \(\overrightarrow{\mathrm{B}}\), so charge particle will move in circular path, whose radius is given by
\(\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}\)
Starting point \(\rightarrow \mathrm{A}\)
Ending point \(\rightarrow \mathrm{C}\)
\(\therefore\) Net displacement \(=\mathrm{AC}\)
\(\begin{aligned} & \mathrm{AC}=\mathrm{CD}-\mathrm{AD} \\ & \mathrm{AC}=\frac{2 \mathrm{mv}}{\mathrm{qB}_1}-\frac{2 \mathrm{mv}}{\mathrm{qB}_2} \\ & \mathrm{AC}=\frac{2 \mathrm{mv}}{\mathrm{qB}_1}\left[1-\frac{\mathrm{B}_1}{\mathrm{~B}_2}\right]\end{aligned}\)
\(\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}\)
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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