In the $xy$-plane, the region $y > 0$ has a uniform magnetic field $B_1\hat{k}$ and the region $y < 0$ has…
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 ________.
Solution
Average speed along the $x$-axis
$V_x = \frac{\int |V_x| dt}{\int dt} = \frac{d_1+d_2}{t_1+t_2} \rightarrow (1)$
We also have,
$\begin{aligned}
&r_1=\frac{mv}{qB_1}, r_2=\frac{mv}{qB_2} \\
&\text{since } B_1=\frac{B_2}{4}
\end{aligned}$
$\therefore r_1=4 r_2 \rightarrow (2)$
Time in $B_1 \Rightarrow \frac{\pi B}{q B_1}=t_1$
Time in $B_2 \Rightarrow \frac{\pi B}{qB_2}=t_2$
Total distance along $x$-axis $d_1+d_2=2 r_1+2 r_2=2(r_1+r_2)=2(5 r_2)$
Total time $T=t_1+t_2=5 t_2$
$\therefore \text{Average speed}=\frac{10 r_2}{5 t_2}$=2 $\frac{mv}{qB_2} \times \frac{qB_2}{\pi m}=2$