Question

A tightly wound long solenoid has $n$ turns per unit length, radius $R$ and carries current $i$. A particle having charge $q$ and mass $m$ is projected from a point on the axis in the direction perpendicular to the axis. The maximum speed for which the particle does not strike the solenoid will be:

• A. $\frac{{\mu }_{0}qRni}{2m}$
• B. $\frac{{\mu }_{0}qRni}{m}$
• C. $\frac{{\mu }_{0}qRni}{3m}$
• D. $\frac{{\mu }_{0}qRni}{4m}$

Answer 1

The correct option is D $\frac{{\mu }_{0}qRni}{4m}$

Let us represent the given scenario (assuming clockwise direction of current) as shown in figure below, the cross-sectional view of solenoid is


From right-hand thumb rule, the magnetic force acting on charge will be towards left ($\text{-ve}x-$direction$)$

The magnetic force on the moving charge in a magnetic field is given by,

${\overrightarrow{F}}_m=q(\overrightarrow{V}\times \overrightarrow{B})$

Here, $[\overrightarrow{V}=V\hat{j};\overrightarrow{B}=B(-\hat{k}\left)\right]$

Thus, the direction of ${\overrightarrow{F}}_m$ is along $[\hat{j}\times (-\hat{k})=-\hat{i}]$.

Charge, so as not to strike the solenoid;

$2r_{max}=\frac{R}{2}$

$2\times \frac{m{V}_{max}}{qB}=\frac{R}{2}(\because r=mV/qB)$

$\frac{m{V}_{max}}{q\left({\mu }_{0}ni\right)}=\frac{R}{4}(\because B={\mu }_{0}ni)$

$\therefore V_{max}=\frac{{\mu }_{0}qniR}{4m}$

Therefore, option $\left(D\right)$ is the correct answer.