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QJEEM23Phc7d6: JEE Main 2023 Physics - Class 12 JEEM23Phc7d6 Solution: Find the mutual inductance in the arrangement, when a small…

Find the mutual inductance in the arrangement, when a small circular loop of wire of radius ' RR ' is placed inside a large square loop of wire of side LL (L≫R)(L \gg R) . The loops are coplanar and their centres coincide :

JEE Main 2023 (Online) 29th January Morning Shift Physics - Electromagnetic Induction Question 43 English

Multiple Choice Options

  • A M=2μ0RL2M=\frac{\sqrt{2} \mu_{0} R}{L^{2}}
  • B M=22μ0RL2M=\frac{2 \sqrt{2} \mu_{0} R}{L^{2}}
  • C M=22μ0R2LM=\frac{2 \sqrt{2} \mu_{0} R^{2}}{L}
  • D M=2μ0R2LM=\frac{\sqrt{2} \mu_{0} R^{2}}{L}

Step-by-Step Verified Solution

JEE Main 2023 (Online) 29th January Morning Shift Physics - Electromagnetic Induction Question 43 English Explanation
amp;B at centre =μ0i4π(L2)(22)×4amp;=2μ0i2πL×4amp;=(22μ0iπL)\begin{aligned}& B \text { at centre }=\frac{\mu_{0} i}{4 \pi\left(\frac{L}{2}\right)}\left(\frac{2}{\sqrt{2}}\right) \times 4 \\\\&= \frac{\sqrt{2} \mu_{0} i}{2 \pi L} \times 4 \\\\&=\left(\frac{2 \sqrt{2} \mu_{0} i}{\pi L}\right)\end{aligned}

Mutual inductance =B⋅Ai=\frac{B \cdot A}{i}

=22μ0iπL×πR2i=\frac{2 \sqrt{2} \mu_{0} i}{\pi L} \times \frac{\pi R^{2}}{i}

=(22μ0R2L)=\left(\frac{2 \sqrt{2} \mu_{0} R^{2}}{L}\right)

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