Question

A particle of charge $q$ and mass $m$ is moving with a velocity $-v\hat{i}(v\ne 0)$ towards a large screen placed in the $(Y-Z)$ plane at a distance $d$. If there is a magnetic field $\overrightarrow{B}=B_0\hat{k}$, the maximum value of $v$ for which the particle will not hit the screen is :

• A. $\frac{qd{B}_0}{3m}$
• B. $\frac{2qd{B}_0}{m}$
• C. $\frac{qd{B}_0}{m}$
• D. $\frac{qd{B}_0}{2m}$

Answer 1

The correct option is C $\frac{qd{B}_0}{m}$

In uniform magnetic field particle moves in a circular path, if the radius of the circular path is nearly equal to ${}^{\prime }{d}^{\prime }$, particle will not hit the screen.
as, $d=\frac{mv}{q{B}_0}[\because \frac{m{v}^2}{d}=qv{B}_0]$

Hence, maximum value of $v$ for which the particle will not hit the screen is

$v=\frac{q{B}_{0}d}{m}$