Question

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

• A. Velocity of the particle when it comes out from the magnetic field is $\overrightarrow{v}=vcos{60}^{\circ }\overrightarrow{J}$
• B. Time for which the particle was in magnetic field is $\frac{\pi m}{3qB}$
• C. Distance travelled in magnetic field is $\frac{\pi mv}{2qB}$
• D. magnitude the momentum of the particle when it exits is mV

Answer 1

Answer: (A), (B), (D)

The correct options are
A Velocity of the particle when it comes out from the magnetic field is $\overrightarrow{v}=vcos{60}^{\circ }\overrightarrow{J}$
B

Time for which the particle was in magnetic field is $\frac{\pi m}{3qB}$

D magnitude the momentum of the particle when it exits is mV

$AB=\frac{\pi }{3}r=\frac{\pi mu}{3qB}Timet=\frac{\theta }{\omega }=\frac{\frac{\pi }{3}}{\frac{2\pi }{T}}=\frac{T}{6}=\frac{\pi }{3qB}Distance=vt=\left(\frac{\pi m}{3qB}\right)v=\frac{\pi mv}{3Bq}$