A particle of mass $m$ and charge $q$ is projected with velocity $v$ into a region of uniform magnetic field $B$ directed into the page. The particle enters the region at an angle $\theta$ with the boundary. If the particle exits the region after a time $t$, the width of the magnetic field region is:

Solution

1. The particle follows a circular path with radius $R=\frac{mv}{qB}$.
2. The velocity component perpendicular to the boundary is $v_{\perp }=v\sin \theta$.
3. The angular displacement in time $t$ is $$\begin{array}{rl}\varphi & =\omega t\\ & =\frac{qBt}{m}\end{array}$$.
4. The displacement perpendicular to the boundary is $d=R\sin \theta -R\sin (\theta -\varphi )$ is incorrect; the correct geometry yields $d=R\sin \theta -R\sin (\theta -\frac{qBt}{m})$.
5. Simplifying the geometry, the width is $d=\frac{mv}{qB}\sin \theta (1-\cos (\frac{qBt}{m}))$. Hence the answer is (D).