JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A small circular loop of wire of radius \(a\) is located at the centre of a much larger circular wire loop of radius \(b\). The two loops are in the same plane. The outer loop of radius \(b\) carries an alternating current \(I = I_0\, cos\, (\omega t)\) . The emf induced in the smaller inner loop is nearly
- A \(\frac{{\pi {\mu _0}{I_0}}}{2}.\frac{{{a^2}}}{b}\omega \,\sin \,\left( {\omega t} \right)\)
- B \(\frac{{\pi {\mu _0}{I_0}}}{2}.\frac{{{a^2}}}{b}\omega \,\cos \,\left( {\omega t} \right)\)
- C \(\pi {\mu _0}{I_0}\,\frac{{{a^2}}}{b}\omega \,\sin \,\left( {\omega t} \right)\)
- D \(\frac{{\pi {\mu _0}{I_0}{b^2}}}{a}\,\omega \,\cos \,\left( {\omega t} \right)\)
Answer & Solution
Correct Answer
(A) \(\frac{{\pi {\mu _0}{I_0}}}{2}.\frac{{{a^2}}}{b}\omega \,\sin \,\left( {\omega t} \right)\)
Step-by-step Solution
Detailed explanation
For two concentric circular coil, Mutual Inductance \(M=\frac{\mu_{0} \pi \mathrm{N}_{1} \mathrm{N}_{2} \mathrm{a}^{2}}{2 \mathrm{b}}\) here, \(\mathrm{N}_{1}=\mathrm{N}_{2}=1\) Hence, \(M=\frac{\mu_{0} \pi a^{2}}{2 b}.........(i)\) and given…
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