Question:
The magnetic field at a distance 'r' from a long straight wire carrying current I is 0.4 tesla. The magnetic field at a distance 2r will be
The magnetic field at a distance 'r' from a long straight wire carrying current I is 0.4 tesla. The magnetic field at a distance 2r will be
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The period is 2 pi m over q B, which has no velocity in it.
Updated On: Oct 1, 2026
- \(0\cdot 1\) tesla
- \(0\cdot 2\) tesla
- \(0\cdot 8\) tesla
- \(1\cdot 6\) tesla
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
A charge moving perpendicular to a field follows a circle. The magnetic force provides the centripetal force: \(qvB = \frac{mv^2}{r}\), so \(r = \frac{mv}{qB}\).
Step 2: Period:
\[ T = \frac{2\pi r}{v} = \frac{2\pi m}{qB} \]
The velocity cancels out. The period depends on mass, charge and the field, but not on the speed.
Final Answer:
The period is independent of the velocity of the particle, option (B). \[ \boxed{\text{velocity of the particle}} \]
A charge moving perpendicular to a field follows a circle. The magnetic force provides the centripetal force: \(qvB = \frac{mv^2}{r}\), so \(r = \frac{mv}{qB}\).
Step 2: Period:
\[ T = \frac{2\pi r}{v} = \frac{2\pi m}{qB} \]
The velocity cancels out. The period depends on mass, charge and the field, but not on the speed.
Final Answer:
The period is independent of the velocity of the particle, option (B). \[ \boxed{\text{velocity of the particle}} \]
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