In the -plane, the region has a uniform magnetic field and the region has another uniform magnetic field A positively charged particle is projected from the origin along the positive -axis with speed at , as shown in the figure. [2018]

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 $v_{0} = π {ms}^{- 1}$ 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 ${ms}^{- 1}$ , along the $x$ -axis in the time interval $T$ is _______. [2018]

Smart Achievers · Accepted Answer

(2)

$Average speed along x -axis,$ $v_{x} = \frac{d_{1} + d_{2}}{t_{1} + t_{2}}$

Here, $r_{1} = \frac{m v_{0}}{q B_{1}}$ and $r_{2} = \frac{m v_{0}}{q B_{2}}$

$∵$ $B_{1} = \frac{B_{2}}{4}$ $∴$ $r_{1} = 4 r_{2}$

Time spent by charged particle in $B_{1}$ , $t_{1} = \frac{π m}{q B_{1}}$

Time spent by charged particle in $B_{2}$ , $t_{2} = \frac{π m}{q B_{2}}$

Total distance along $x$ -axis,

$d_{1} + d_{2} = 2 r_{1} + 2 r_{2} = 2 (r_{1} + r_{2}) = 2 (5 r_{2}) = 10 r_{2}$

$Average speed = \frac{10 r_{2}}{5 t_{2}} = 2 \frac{m v_{0}}{q B_{2}} \times \frac{q B_{2}}{π m}$ $= \frac{2 v_{0}}{π}$

$= 2 {ms}^{- 1}$ $(∵ v_{0} = π {ms}^{- 1})$

Solution

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