In the xy-plane, the region $y > 0$ has a uniform field $B_{1} k$ and the region $y < 0$ has another uniform magnetic field $B_{2} k$ . A positively charged particle is projected from the origin along positive y-axis with speed $υ_{0} = {πs}^{- 1}$ 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 _______

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Detailed Solution

(i) Average speed along x-axis

$(u_{x}) = \frac{\int | \vec{u_{x}} | d x}{\int d x} = \frac{d_{1} + d_{2}}{t_{1} + t_{2}}$
(ii) We have,
$r_{1} = \frac{m υ}{q B_{1}}, r_{2} = \frac{m υ}{q B_{2}}$
Since, $B_{1} = \frac{B_{2}}{4}$
$∴ r_{1} = 4 r_{2}$
Time in $B_{1} \Rightarrow \frac{π m}{q B_{1}} = t_{1}$
Time in $B_{2} \Rightarrow \frac{π m}{q B_{2}} = t_{2}$
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 $T = t_{1} + t_{2} = 5 t_{2}$
Average speed $= \frac{10 r_{2}}{5 t_{2}} = \frac{2 m v}{q B_{2}} \times \frac{q B_{2}}{π m} = 2 m s^{- 1}$