Question:
A charge moves in a circular path perpendicular to a magnetic field. The time period of revolution is independent of
A charge moves in a circular path perpendicular to a magnetic field. The time period of revolution is independent of
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Period T = 2 pi m / (q B), independent of speed and radius.
Updated On: Oct 1, 2026
- mass of the particle
- velocity of the particle
- magnetic field
- charge
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
A charge moving perpendicular to a magnetic field follows a circle. The magnetic force supplies the centripetal force: \(qvB = \dfrac{mv^2}{r}\), so \(r = \dfrac{mv}{qB}\).
Step 2: Period
\[ T = \frac{2\pi r}{v} = \frac{2\pi m}{qB} \]
The period depends on the mass, the charge and the field, but the speed v cancels out. So the period is independent of the velocity of the particle, option (B).
Final Answer:
The period is independent of the velocity, option (B). \[ \boxed{T = \frac{2\pi m}{qB}} \]
A charge moving perpendicular to a magnetic field follows a circle. The magnetic force supplies the centripetal force: \(qvB = \dfrac{mv^2}{r}\), so \(r = \dfrac{mv}{qB}\).
Step 2: Period
\[ T = \frac{2\pi r}{v} = \frac{2\pi m}{qB} \]
The period depends on the mass, the charge and the field, but the speed v cancels out. So the period is independent of the velocity of the particle, option (B).
Final Answer:
The period is independent of the velocity, option (B). \[ \boxed{T = \frac{2\pi m}{qB}} \]
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