Question

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

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

Answer 1

The correct option is A

$\frac{qd{B}_0}{m}$

Step 1: Given data.

Mass of the particle$=m$

Particle of charge$=q$

Velocity $=v\hat{i}$

The magnetic field strength of the particle $B=B_0\hat{k}$

Step 2: Finding the minimum velocity $v_{min}$ for which the particle will not hit the screen.

As we know in case of force on a charged particle in a magnetic field, the cyclotron equation is given as:

$R=\frac{mv}{q{B}_0}$ …….$\left(i\right)$

Where $R$ is the radius of the circular path and $v$ is the velocity.

Since, if the charged particle should not hit the screen then the radius of the circular path is less than distance $d$ such as.

$R\le d$ $\left[\mathbf{\because }\mathit{R}\mathbf{=}\frac{\mathbf{m}\mathbf{v}}{\mathbf{q}{\mathbf{B}}_{\mathbf{0}}}\right]$

$\Rightarrow$$\frac{mv}{q{B}_0}\le d$ $\left[\because fromequation\left(i\right)\right]$

$\Rightarrow$ $v\le \frac{qd{B}_0}{m}$

Thus, the maximum value of $\mathit{v}\mathbf{=}\frac{\mathbf{q}\mathbf{d}{\mathbf{B}}_{\mathbf{0}}}{\mathbf{m}}$

Hence, option A is correct.