Question

In region $x>0$ , a uniform and constant magnetic field $\overrightarrow{B_1}=2B_0\overrightarrow{k}$ exists. Another uniform and constant magnetic field $\overrightarrow{B_2}=B_0\hat{k}$, exists in region $x<0$. A positively charged particle of mass $m$ and charge $q$ is crossing origin at time $t=0$ with a velocity $\overrightarrow{u}=u_0\hat{i}$. The particle comes back to its initial position after a time $(B_0,u_0$ are positive constants)

• A. $\frac{3}{2}\frac{\pi m}{q{B}_0}$
• B. $\frac{2\pi m}{q{B}_0}$
  
• C. $\frac{3\pi m}{q{B}_0}$
• D. The particle does not come back to its initial position.

Answer 1

The correct option is B $\frac{2\pi m}{q{B}_0}$



It will make a semicircular path in $x>0$. So, time to go from point $O$ to $P$,

$t_1=\frac{T}{2}=\frac{2\pi m}{2qB}=\frac{\pi m}{q\left(2B_0\right)}$

And radius, $r_1=\frac{mv}{qB}=\frac{m{u}_0}{q\left(2B_0\right)}$

Now, when it will reach $x<0$ region its radius will become

$r_2=\frac{mv}{qB}=\frac{mv}{q{B}_0}$

So, $r_2=2r_1$.

Time to go from $P$ to $Q$

$t_2=\frac{T^{\prime }}{2}=\frac{2\pi m}{qB\times 2}=\frac{\pi m}{q{B}_0}$

Now, it will enter in $x>0$ region.

Time required to go from $Q$ to $O$ will be same as $O$ to $P$.
So, total time to reach origin $O$.

$T_t=t_1+t_2+t_3=t_1+t_2+t_1$

$\Rightarrow T_t=\frac{\pi m}{2q{B}_0}+\frac{\pi m}{q{B}_0}+\frac{\pi m}{2q{B}_0}$

$\therefore T_t=\frac{2\pi m}{q{B}_0}$

Hence, option (b) is the correct answer.