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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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 \(\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,N/{A^2}} \right)\)
- A \(2 \times {10^{ - 6}}\,C\)
- B \(7 \times {10^{ - 6}}\,C\)
- C \(4 \times {10^{ - 5}}\,C\)
- D \(3 \times {10^{ - 5}}\,C\)
Answer & Solution
Correct Answer
(D) \(3 \times {10^{ - 5}}\,C\)
Step-by-step Solution
Detailed explanation
\(B=\frac{\mu_{0} i}{2 R}=\frac{\mu_{0} q \omega}{2 R 2 \pi}\) \(\Rightarrow q=3 \times 10^{-5}\, \mathrm{C}\)
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