A positive charge 'q' of mass 'm' is moving along the + x axis. We wish to apply a uniform magnetic field B for time Δ t so that the charge reverses its direction crossing the y axis at a distance d. Then

(A) B\, = \,\frac{{mv}}{{qd}} and \Delta t\, = \,\frac{{\pi d}}{v}

(B) B\, = \,\frac{{mv}}{{2qd}} and \Delta t\, = \,\frac{{\pi d}}{2v}

(D) B\, = \,\frac{{2mv}}{{qd}} and \Delta t\, = \,\frac{{\pi d}}{v}

JEE Mains · Physics · STD 12 - Moving charges and magnetism

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

Pick an option to check your answer.

Verified Solution

Correct Answer

(C) \(B\, = \,\frac{{2mv}}{{qd}}\) and \(\Delta t\, = \,\frac{{\pi d}}{2v}\)

Detailed explanation

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).

\(\Delta \mathrm{t}=\frac{\pi \mathrm{d}}{2 \mathrm{v}}\)

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