Question

In the x-y-plane, the region $y>0$ has a uniform magnetic field $B_1\hat{k}$ and the region $y<0$ has another uniform magnetic field $B_2\hat{k}$. A positively charged particle is projected from the origin along the positive $y$-axis with speed $V_o=\pi m{s}^{-1}$ at $t=0$, as shown in the figure. Neglect gravity in this problem. Let $t=T$ be the time when the particle crosses the x-axis from below for the first time. If $B_2=4B_1$, the average speed of the particle, in $m{s}^{-1}$, along the x-axis in the time interval $T$ is

Answer 1

Average speed along the x-axis
$\begin{array}{}\left(V_x\right)=\frac{\int |{\overrightarrow{V}}_x|dt}{\int dt}=\frac{d_1+d_2}{t_1+t_2}& \text{(1)}\end{array}$
We also have,
$r_1=\frac{m{V}_o}{q{B}_1},r_2=\frac{m{V}_o}{q{B}_2}$
since $B_1=\frac{B_2}{4}$
$\therefore r_1=4r_2$ $\to$ $\left(2\right)$
Time in $B_1\Rightarrow t_1=\frac{\pi m}{q{B}_1}$
Time in $B_2\Rightarrow t_2=\frac{\pi m}{q{B}_2}$
Total distance along x-axis $d_1+d_2=2r_1+2r_2=2(r_1+r_2)$$=2\left(5r_2\right)$
Total time $T=t_1+t_2=5t_2$
$\therefore$ Average speed $=\frac{10r_2}{5t_2}=2\frac{m{V}_o}{q{B}_2}\times \frac{q{B}_2}{\pi m}=2$ $(\because V_o=\pi m/s)$