Question:easy
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
Show Hint
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
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Radius
$r = mv/(qB)$ grows with $v$.
Step 2: Time for one turn
Distance $2\pi r$ over speed $v$ gives $2\pi m/(qB)$: $v$ cancels.
Step 3: Pick
Velocity, option (B).
Final Answer:
Option (B). \[ \boxed{T = \frac{2\pi m}{qB}} \]
$r = mv/(qB)$ grows with $v$.
Step 2: Time for one turn
Distance $2\pi r$ over speed $v$ gives $2\pi m/(qB)$: $v$ cancels.
Step 3: Pick
Velocity, option (B).
Final Answer:
Option (B). \[ \boxed{T = \frac{2\pi m}{qB}} \]
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