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Class 12 › Physics › Moving Charges and Magnetism
CBSE2022Class Class 12 · Physics3 Marks · Short✅ Verified
A proton is moving with speed $2 \times 10^5$ m/s and enters a uniform magnetic field $B = 1.5$ T. The velocity makes an angle of 30° with $\vec{B}$. Calculate: (a) the pitch of the helical path, (b) kinetic energy after completing half the circle.
✅ Answer & Solution
(a) Pitch of helical path:
Component of velocity along $B$: $v_{||} = v\cos 30° = 2 \times 10^5 \times \frac{\sqrt{3}}{2} = \sqrt{3} \times 10^5$ m/s
Component perpendicular to $B$: $v_{\perp} = v\sin 30° = 10^5$ m/s
Time period: $T = \frac{2\pi m}{qB} = \frac{2\pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 1.5} = 4.37 \times 10^{-8}$ s
$$\text{Pitch} = v_{||} \times T = \sqrt{3} \times 10^5 \times 4.37 \times 10^{-8} = 7.57 \times 10^{-3} \text{ m}$$
(b) Kinetic energy:
Magnetic force does no work, so KE remains constant:
$$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1.67 \times 10^{-27} \times (2 \times 10^5)^2 = 3.34 \times 10^{-17} \text{ J}$$
