Question

A charge particle of charge q and mass m is moving with velocity v as shown in fig in a uniform magnetic field B along -ve z-direction. Select the correct alternative(s)

• A. velocity of the particle when it comes out from the magnetic field is $\overrightarrow{v}=v(\cos {60}^0\hat{i}+\sin {60}^0\hat{j})$
• B. time for which the particle was in magnetic field is $\frac{\pi m}{3qB}$
• C. distance travelled in magnetic field is $\frac{\pi mv}{3qB}$
• D. none of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A velocity of the particle when it comes out from the magnetic field is $\overrightarrow{v}=v(\cos {60}^0\hat{i}+\sin {60}^0\hat{j})$
B time for which the particle was in magnetic field is $\frac{\pi m}{3qB}$
C distance travelled in magnetic field is $\frac{\pi mv}{3qB}$

Arc length $AB=\frac{\pi }{3}r=\frac{\pi }{\frac{mv}{qB}}=\frac{\pi qB}{mv}$

Time spent inside magnetic field $t=\frac{\frac{\pi }{3}}{\frac{2\pi }{\frac{2\pi m}{qB}}}=\frac{\pi m}{3qB}$

Distance traveled inside magnetic field $s=v\frac{\pi mv}{3qB}=\frac{\pi m}{3qB}$

Velocity vector $\overrightarrow{v}$ makes an ${60}^0$ with x-axis when it come out of the magnetic firld.

$\therefore \overrightarrow{v}=v(\cos {60}^0\hat{i}+\sin {60}^0\hat{j})$