Question

In the $xy$ - 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 $v_0=\pi m{s}^{-1}$ 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

Answer: 2

(i) Average speed along $x$ - axis
$⟨v_x⟩=\frac{\int |{\vec{v}}_x|dt}{\int dt}=\frac{d_1+d_2}{t_1+t_2}$
(ii) We have,
$r_1=\frac{mv}{qB},r_2=\frac{mv}{q{B}_2}$
Since, $B_1=\frac{B_2}{4}$
$\therefore r_1=4r_2$
Time in $B_2\Rightarrow \frac{\pi m}{q{B}_2}=t_2$
Total distance along $x$ - axis $d_1+d_2=2r_1+2r_2=2\left(r_1+r_2\right)=2$
$\left(5r_2\right)$
Total time $T=t_1+t_2=5t_2$
$\therefore$ Average speed $=\frac{10r_2}{5t_2}=\frac{2mv}{q{B}_2}\times \frac{a{B}_2}{q{B}_2}=2$