Question:
A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of \(40\pi\) rad s\(^{-1}\) about its axis, perpendicular to its plane. If the magnetic field at its centre is \(3.8 \times 10^{-9}\) T, then the charge carried by the ring is close to (\(\mu_0 = 4\pi \times 10^{-7}\) N/A\(^2\)).
A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of \(40\pi\) rad s\(^{-1}\) about its axis, perpendicular to its plane. If the magnetic field at its centre is \(3.8 \times 10^{-9}\) T, then the charge carried by the ring is close to (\(\mu_0 = 4\pi \times 10^{-7}\) N/A\(^2\)).
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A rotating charged ring is equivalent to a current loop with \(I = Q\omega/2\pi\); use \(B = \mu_0 I/2R\) at the centre.
Updated On: Nov 5, 2026
- \(7 \times 10^{-6}\) C
- \(4 \times 10^{-5}\) C
- \(2 \times 10^{-6}\) C
- \(3 \times 10^{-5}\) C
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The Correct Option is D
Solution and Explanation
A uniformly charged ring of charge \(Q\) rotating with angular speed \(\omega\) behaves as a circular current loop.
Current: \(I = \dfrac{Q}{T} = \dfrac{Q\omega}{2\pi}\).
Field at the centre: \(B = \dfrac{\mu_0 I}{2R} = \dfrac{\mu_0 Q\omega}{4\pi R}\).
So \(Q = \dfrac{4\pi R B}{\mu_0 \omega}\).
Substituting \(R = 0.1\) m, \(B = 3.8\times10^{-9}\) T, \(\mu_0 = 4\pi\times10^{-7}\), \(\omega = 40\pi\):
\(Q = \dfrac{4\pi \times 0.1 \times 3.8\times10^{-9}}{4\pi\times10^{-7}\times 40\pi} = \dfrac{3.8\times10^{-10}}{40\pi\times10^{-7}} = \dfrac{3.8\times10^{-10}}{1.2566\times10^{-5}} \approx 3.0\times10^{-5}\) C.
Answer: (4) \(3 \times 10^{-5}\) C
Current: \(I = \dfrac{Q}{T} = \dfrac{Q\omega}{2\pi}\).
Field at the centre: \(B = \dfrac{\mu_0 I}{2R} = \dfrac{\mu_0 Q\omega}{4\pi R}\).
So \(Q = \dfrac{4\pi R B}{\mu_0 \omega}\).
Substituting \(R = 0.1\) m, \(B = 3.8\times10^{-9}\) T, \(\mu_0 = 4\pi\times10^{-7}\), \(\omega = 40\pi\):
\(Q = \dfrac{4\pi \times 0.1 \times 3.8\times10^{-9}}{4\pi\times10^{-7}\times 40\pi} = \dfrac{3.8\times10^{-10}}{40\pi\times10^{-7}} = \dfrac{3.8\times10^{-10}}{1.2566\times10^{-5}} \approx 3.0\times10^{-5}\) C.
Answer: (4) \(3 \times 10^{-5}\) C
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