JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A rod with circular cross-section area \(2\,cm ^2\) and length \(40\,cm\) is wound uniformly with \(400\) turns of an insulated wire. If a current of \(0.4\,A\) flows in the wire windings, the total magnetic flux produced inside windings is \(4\,\pi \times 10^{-6}\,Wb\). The relative permeability of the rod is (Given : Permeability of vacuum \(\left.\mu_0=4 \pi \times 10^{-7} \;NA ^{-2}\right)\)
- A \(12.5\)
- B \(\frac{32}{5}\)
- C \(125\)
- D \(\frac{5}{16}\)
Answer & Solution
Correct Answer
(C) \(125\)
Step-by-step Solution
Detailed explanation
\(\phi=\mu_{ r } \mu_{ o } \frac{ N }{\ell} I \times A\) \(\mu_{ r }=125\)
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