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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 $\Delta t$ so that the charge reverses its direction crossing the $y$ axis at a distance $d.$ Then
$B\, = \,\frac{{mv}}{{qd}}$ and $\Delta t\, = \,\frac{{\pi d}}{v}$
$B\, = \,\frac{{mv}}{{2qd}}$ and $\Delta t\, = \,\frac{{\pi d}}{2v}$
$B\, = \,\frac{{2mv}}{{qd}}$ and $\Delta t\, = \,\frac{{\pi d}}{2v}$
$B\, = \,\frac{{2mv}}{{qd}}$ and $\Delta t\, = \,\frac{{\pi d}}{v}$
Solution
The applied magnetic field provides the required centripetal force to the charge particle, so it can move in circular path of radius $\frac{d}{2}$
$\therefore \mathrm{Bqv}=\frac{\mathrm{mv}^{2}}{\mathrm{d} / 2}$
or, $B = \frac{{2{\text{mv}}}}{{{\text{qd}}}}$
Time interval for which a uniform magnetic field is applied
$\Delta t=\frac{\pi \cdot \frac{d}{2}}{v}$
(particle reverses its direction after time $\Delta t$ by covering semi circle).
$\Delta \mathrm{t}=\frac{\pi \mathrm{d}}{2 \mathrm{v}}$
