Question

A particle of mass $m$ and charge $q$ moves with constant velocity $v$ along positive $x-$direction. It enters a region containing a uniform magnetic field $B$ directed along the negative $z-$direction, extending from $x=a$ to $x=b$. The minimum value of $v$ required so that the particle can just enter the region $x>b$ is:

• A. $\frac{qbB}{m}$
• B. $\frac{q(b-a)B}{m}$
• C. $\frac{qaB}{m}$
• D. $\frac{q(b+a)B}{2m}$

Answer 1

The correct option is B $\frac{q(b-a)B}{m}$

As shown in figure below, the $\text{-ve}$ $z-$direction is normally inwards to the plane of paper.


As shown in the above figure, at the limiting value of velocity $v$, the particle will describe a circular path by just touching point $\text{P}$.

If $v>v_{min}$, the particle will enter region $x>b$ through point $P$.

For limiting case, the radius of the circular path,
$R=b-a$

$\frac{m{v}_{min}}{qB}=b-a(\because R=mv/qB)$

$v_{min}=\frac{qB(b-a)}{m}$

Hence, option $\left(b\right)$ is the correct answer.

Why this question ? Tip: In problems involving magnetic field in certain region, the limiting condition of velocity to exit the region will correspond to the value of velocity "when the charged particle just describes the circle by touching the outer edge of magnetic field region".