JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A small square loop of wire of side \(l\) is placed inside a large square loop of wire \(L(L \gg l)\). Both loops are coplanar and their centres coincide at point \(O\) as shown in figure. The mutual inductance of the system is.
- A \(\frac{2 \sqrt{2} \mu_{0} L ^{2}}{\pi \ell}\)
- B \(\frac{\mu_{0} \ell^{2}}{2 \sqrt{2 \pi L}}\)
- C \(\frac{2 \sqrt{2} \mu_{0} \ell^{2}}{\pi L }\)
- D \(\frac{\mu_{0} L ^{2}}{2 \sqrt{2} \pi \ell}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 \sqrt{2} \mu_{0} \ell^{2}}{\pi L }\)
Step-by-step Solution
Detailed explanation
Assuming current I in outer loop magnetic field at \(\text { centre }=4 \times \frac{\mu_{0} i}{4 \pi \times \frac{L}{2}} \times\left(2 \sin 45^{\circ}\right)=\frac{2 \sqrt{2} \mu_{0} i}{\pi L }\) \(M =\frac{\text { Flux through inner loop }}{ i }\)…
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