Question

A positively charged particle having charge $q$ and mass $m$ moving with velocity $v_0\hat{i}$ enters a region in which magnetic field $\overrightarrow{B}=-B_0\hat{k}$ exists. The magnetic field region extends from $x=0$ to $x=d$. Then the time spent by the particle in magnetic field if $d$ is equal to $\frac{1.5mv}{qB}$

• A. $\frac{\pi m}{qB}$
• B. $\frac{2\pi m}{qB}$
• C. $\frac{2\pi m}{3qB}$
• D. $\frac{\pi m}{4qB}$

Answer 1

The correct option is A $\frac{\pi m}{qB}$

The radius of the circular path of charge particle is, $R=\frac{mv}{qB}$

Here width of field region is;

$d=\frac{1.5mv}{qB}=1.5R$

$\Rightarrow d>R$

$\because d>R$ hence particle will complete its path in magnetic field and won't emerge out from the opposite boundary.

Considering the symmetry of motion at entry and emergence;


The path is a semicircle of radius $R$

Time spent in magnetic field will be,

$t=\frac{T}{2}$
$\Rightarrow t=\frac{\left(\frac{2\pi m}{qB}\right)}{2}$

$\therefore t=\frac{\pi m}{qB}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (a) is the correct answer.

Why this question ? Tip:If we draw the magnetic force $\overrightarrow{F_m}$ by using $\overrightarrow{F_m}=q(\overrightarrow{v}\times \overrightarrow{B})$ at entry and emergence, then the inter section point of forces will give centre of circular path.