Question

A particle of charge -q and mass m enters a uniform magnetic field $\overrightarrow{B}$ at A with speed $v_1$ at an angle $\alpha$ and leaves the field at C with speed $v_2$ at an angle $\beta$ as shown. Then:

• A. $\alpha =\beta$
• B. $v_1=v_2$
• C. Particle remains in the field for time $t=\frac{2m(\pi -\alpha )}{qB}$
• D. All of these

Answer 1

The correct option is C Particle remains in the field for time $t=\frac{2m(\pi -\alpha )}{qB}$

Since the particle is of negative charge it will move in circular do it as shown.

The centripetal force $f$ will be equal to Lorentz force at any point

i.e $F=qvB$

$\frac{n{v}^2}{r}=qvB$

$v=\frac{qrB}{m}$

Since here $q,r,B$ and $m$ are constant quantity therefore velocity will be constant in whole path

$\Rightarrow v_1=v_2=v=\frac{qrB}{m}$

$\angle MAO=180-(90+\alpha )$

$=90-\alpha$

$\angle MAO=90-\beta$

Since the $△AMO$ and $△BMO$ is similar by $RHS$ property of similarity

$\Rightarrow \angle MAO=\angle MBO$

$i.e90-\alpha =90-\beta$

$\Rightarrow \alpha =\beta$

Now,

In $△AOB$

$\angle A+\angle O+\angle B={180}^o$

$90-\alpha +\alpha O+90-\alpha ={180}^o$

$\Rightarrow \angle O+180-2\alpha ={180}^o$

$\Rightarrow \angle O=\angle AOB=2\alpha$

We know that

$\theta =\frac{l}{r}$

$\Rightarrow l=\theta r$

distance travelled by particle $=(2\pi -2\alpha )r$

$\Rightarrow$ Time taken $=\frac{distancetravelled}{speed}$

$=\frac{2(\pi -\alpha )r}{qrB/m}$

$\Rightarrow$ Time taken $=\frac{2m(\pi -\alpha }{qB}$