Question

The region between \( y = 0 \) and \( y = d \) contains a magnetic field \( \overrightarrow{B} = B\hat{z} \). A particle of mass m and charge q enters the region with a velocity \( \overrightarrow{v} = v\hat{i} \). If \( d = \frac{mv}{2qB} \), the acceleration of the charged particle at the point of its emergence at the other side is :

• A. \( \frac{qvB}{m}\left(-\frac{\sqrt{3}}{2}\hat{i} - \frac{1}{2}\hat{j}\right) \)
• B. \( \frac{qvB}{m}\left(\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}\right) \)
• C. \( \frac{qvB}{m}\left(\frac{-\hat{j} + \hat{i}}{\sqrt{2}}\right) \)
• D. \( \frac{qvB}{m}\left(\frac{\hat{j} + \hat{i}}{\sqrt{2}}\right) \)

Hint:

The radius is \( r = \frac{mv}{qB} = 2d \); the particle turns through an angle \( \theta \) with \( d = r(1-\cos\theta) \), and the acceleration is perpendicular to the velocity, of magnitude \( \frac{qvB}{m} \).

Answer 1

The Correct Option is A

Radius of the circular path \( r = \frac{mv}{qB} \), so \( d = \frac{r}{2} \).
The particle crosses the other boundary when its sideways displacement is d: \( r(1-\cos\theta) = \frac{r}{2} \Rightarrow \cos\theta = \frac12 \Rightarrow \theta = 60^\circ \).
The force on the (positive) charge is \( q\vec{v}\times\vec{B} = qvB(\hat{i}\times\hat{z}) = -qvB\,\hat{j} \), so the path bends towards \( -\hat{j} \) (the geometry is mirrored about the line of entry).
After turning by \( 60^\circ \) the velocity is \( v\left(\cos60^\circ\hat{i} - \sin60^\circ\hat{j}\right) \), and the acceleration is perpendicular to it with magnitude \( \frac{qvB}{m} \):
\( \vec{a} = \frac{qvB}{m}\left(-\sin60^\circ\hat{i} - \cos60^\circ\hat{j}\right) = \frac{qvB}{m}\left(-\frac{\sqrt3}{2}\hat{i} - \frac12\hat{j}\right) \).
Answer: (1) \( \frac{qvB}{m}\left(-\frac{\sqrt{3}}{2}\hat{i} - \frac{1}{2}\hat{j}\right) \)