A positive charge -q- of mass -m- is moving along the -x axis- We wish to apply a uniform magnetic field B time Delta t- so that the charge reverses its direction- crossing the y axis a distance d- Then-displaystyle B-dfrac -mv-qd- and displaystyle Delta t-dfrac -pi d-v-displaystyle B-dfrac -mv-2qd- and displaystyle Delta t-dfrac -pi d-2v-displaystyle B-dfrac -2mv-qd- and displaystyle Delta t-dfrac -pi d-2v-displaystyle B-dfrac -2mv-qd- and displaystyle Delta t-dfrac -pi d-v-

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Toppr Admin · Accepted Answer

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A positive charge 'q' of mass 'm' is moving along the +x axis. We wish to apply a uniform magnetic field B for time Δt, so that the charge reverses its direction, crossing the y axis at a distance d. Then:B=mvqd and Δt=πdvB=mv2qd and Δt=πd2vB=2mvqd and Δt=πd2vB=2mvqd and Δt=πdv

To change the direction of charge, a force needs to be applied such that,qvB=mv2r where r=d/2Therefore,B=2mvqdTotal displacement is d when charge travels for time, δtTherefore,δt=πd2v

To change the direction of charge, a force needs to be applied such that,
$q v B = \frac{m v^{2}}{r}$ where $r = d / 2$
Therefore,
$B = \frac{2 m v}{q d}$
Total displacement is $d$ when charge travels for time, $δ t$
Therefore,
$δ t = \frac{π d}{2 v}$