Question

A particle moving with velocity $v$ having specific charge $\left(\frac{q}{m}\right)$ enters a region of magnetic field $B$ having width $d=\frac{3mv}{5qB}$ at angle ${53}^o$ to the boundary of magnetic field. Find the angle $\theta$ in the shown figure.

• A. ${37}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. None

Answer 1

The correct option is C ${90}^{\circ }$

From the diagram shown, we can see that angle ${53}^{\circ }$ at entry and angle $\theta$ at emergence is from vertical.

Angle of deflection $=\theta -{53}^{\circ }$

${\theta }_d=\theta -{53}^{\circ }$

Using the relation for angle of deflection, $\sin {\theta }_d=\frac{d}{R}$ we have,

$\Rightarrow \sin (\theta -{53}^{\circ })=\frac{\left(\frac{3mv}{5qB}\right)}{\left(\frac{mv}{qB}\right)}$

$\Rightarrow \sin (\theta -{53}^{\circ })=\frac{3}{5}$

$[\because \sin {37}^{\circ }=\frac{3}{5}]$

$\Rightarrow \theta -{53}^{\circ }={37}^{\circ }$

$\therefore \theta ={53}^{\circ }+{37}^{\circ }={90}^{\circ }$

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