In the x-y-plane- the region y -0 has a uniform magnetic field B 1k- and the region y -0 has another uniform magnetic field B 2k- A positively charged particle is projected from the origin along the positive y-axis with speed V 0- 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

Average speed along the x axis (V_x) = ∫ |V_x|dt∫ dt = d_1+d_2t_1+t_2 1 nbsp; We also have, r_1 = mV_oqB_1, r_2 = mV_oqB_2 since B_1=B_24 ∴ r_1=4r_2 → (2) Tim ...

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Standard XII

Physics

Introducing Flux

In the x-y-pl...

Question

In the x-y-plane, the region $y > 0$ has a uniform magnetic 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 the positive $y$ -axis with speed $V_{o} = π m s^{- 1}$ 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

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Solution

Average speed along the x-axis
$\begin{matrix} (V_{x}) = \frac{\int | {\to V}_{x} | d t}{\int d t} = \frac{d_{1} + d_{2}}{t_{1} + t_{2}} \\ (1) \end{matrix}$
We also have,
$r_{1} = \frac{m V_{o}}{q B_{1}}, r_{2} = \frac{m V_{o}}{q B_{2}}$
since $B_{1} = \frac{B_{2}}{4}$
$∴ r_{1} = 4 r_{2}$ $\to$ $(2)$
Time in $B_{1} \Rightarrow t_{1} = \frac{π m}{q B_{1}}$
Time in $B_{2} \Rightarrow t_{2} = \frac{π m}{q B_{2}}$
Total distance along x-axis $d_{1} + d_{2} = 2 r_{1} + 2 r_{2} = 2 (r_{1} + r_{2})$ $= 2 (5 r_{2})$
Total time $T = t_{1} + t_{2} = 5 t_{2}$
$∴$ Average speed $= \frac{10 r_{2}}{5 t_{2}} = 2 \frac{m V_{o}}{q B_{2}} \times \frac{q B_{2}}{π m} = 2$ $(∵ V_{o} = π m / s)$

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