First slide

Force on a charged particle moving in a magnetic field

Question

A proton is projected with kinetic energy 80 Mev in a uniform magnetic field making an angle of $30^{0}$ with the field and pitch of its trajectory is found to be 1m. What will be the pitch of an α-particle if it is projected with same kinetic energy in the same magnetic field making an angle of $60^{o}$ with the field?

Moderate

A

$\frac{\sqrt{3}}{2} m$
B

$\frac{2}{\sqrt{3}} m$
C

$\sqrt{3} m$
D

$\frac{1}{\sqrt{3}} m$

Solution

$\begin{array}{l} \frac{1}{2} {mv}^{2} = k \Rightarrow v = \sqrt{\frac{2 k}{m}} Also, p = vcosθ \times T \\ ∴ p = \sqrt{\frac{2 k}{m}} cosθ \times \frac{2 πm}{QB} = \frac{2 π \sqrt{2 k}}{B} \times \frac{\sqrt{m} . cosθ}{Q} \Rightarrow p \infty \frac{\sqrt{m} . cosθ}{Q} \\ ∴ \frac{p_{α}}{p_{p}} = \sqrt{\frac{m_{α}}{m_{p}}} . \frac{Q_{α}}{Q_{p}} . \frac{\cos 60^{o}}{\cos 30^{o}} = \sqrt{4} . \frac{1}{2} . \frac{1 / 2}{\sqrt{3} / 2} = \frac{1}{\sqrt{3}} \\ \Rightarrow p_{α} = \frac{1}{\sqrt{3}} m \end{array}$

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