A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width 'd'. If ' α ' be the angle of deviation of proton from initial direction of motion (see figure), the value of sin α will be: (a) B/2√(qd/mV) (b) B/d√(q/2mV) (c) Bd√(q/2mV) (d) qV√(Bd/2m)

Answer: (c). After acceleration: 1/2mv² = qV ⇒ mv = √(2mqV). Radius in the field: r = mv/qB = (√(2mqV))/(qB). From the geometry, sin α = d/r = (qBd)/(√(2mqV)) = Bd√(q/2mV) So the answer is (c).

A proton (mass m) accelerated by a potential difference V flies… | JEE Main 2015 (10 Apr, Shift 1)

Q99JEE Main 2015 · 10 APR · SHIFT 1Difficult

Moving Charge in a Magnetic Field

A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width 'd'. If ' $\alpha$ ' be the angle of deviation of proton from initial direction of motion (see figure), the value of sin $\alpha$ will be:

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Answer: (c)

After acceleration: $\tfrac{1}{2}mv^2 = qV \Rightarrow mv = \sqrt{2mqV}$ .
Radius in the field: $r = \dfrac{mv}{qB} = \dfrac{\sqrt{2mqV}}{qB}$ .
From the geometry, $\sin\alpha = \dfrac{d}{r} = \dfrac{qBd}{\sqrt{2mqV}} = Bd\sqrt{\dfrac{q}{2mV}}$
So the answer is (c).

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