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 $B_2=4 B_1$, the average speed of the particle, in $m s ^{-1}$, along the $x$-axis in the time interval $T$ is. . . . . .
IIT 2018, Advanced
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$R _1=\frac{ mv }{ qB }$
$R _2=\frac{ mv }{ qB }$
$B _2=4 B _1$
$R _2=\frac{1}{4} R _1$
Distance traveled in $x$ direction
$\Delta x =2 R _1+2 R _2$
$\Delta x =2 R _1+\frac{ R _1}{2}=\frac{5 R_1}{2}$
$T _1=\frac{\pi m }{ qB }$
$T _2=\frac{\pi m }{ qB }=\frac{ T _1}{4}$
$\text { Total time }=\frac{T_1}{2}+\frac{T_2}{2}=\frac{5 T_1}{8}$
Average speed $V =\frac{\Delta x }{\Delta t }=\frac{\left(\frac{5 R _1}{2}\right)}{\left(\frac{5 T _1}{8}\right)}=4 \frac{ R _1}{ T _1}$
$R=\frac{m V}{q B}$
$T=\frac{2 \pi m}{q B}$
Average speed $=\frac{4 R }{ T }=\frac{4 V }{2 \pi}=2$
Average speed $=2$
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