⇦ | ![]() | ⇨ |
A thin ring of radius R meter has charge q coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of f revolutions/s. The value of magnetic induction in Wb/m² at the centre of the ring is
Options
(a) µ₀q f / 2π R
(b) µ₀q / 2π f R
(c) µ₀q / 2 f R
(d) µ₀q f / 2 R
Correct Answer:
µ₀q f / 2 R
Explanation:
Current in the ring due to the rotation,
I = q/T = q? / 2π = q.2π f / 2π
Therefore, magnetic field at the centre of the ring is
B = µ₀I / 2R = µ₀/2R . q2π f / 2π
= µ₀q f / 2 R
Related Questions:
- A body of mass 4 kg is accelerated upon by a constant force travels
- The lowest frequency of light that will cause the emission of photoelectrons
- Before using the tangent galvanometer, its coil is set in
- A transformer of 100% efficiency has 200 turns in the primary coil and 40000 turns
- A tuned amplifier circuit is used to generate a carrier frequency of 2 MHz
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply