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  • A small square loop of wire of side l is placed inside a large square loop of wire of side L(>> l ). The loops are coplanar and their centres coincide. The mutual inductance of the system is:

A small square loop of wire of side l is placed inside a large square loop of wire of side L(>> l ). The loops are coplanar and their centres coincide. The mutual inductance of the system is:

Option: 1

\mathrm{\text { Directly proportional to } 1^2}


Option: 2

\mathrm{\text { Inversely proportional to } \mathrm{L}^2}


Option: 3

\mathrm{\text { Directly proportional } \mathrm{L}^2}


Option: 4

\mathrm{\text { Inversely proportional to } \mathrm{L}^2}


Answers (1)

best_answer

\mathrm{\text { Because of larger loop, the field at the centre will be } B=\frac{4 \times \mu_0 i}{4 \pi(L / 2)}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)}

\mathrm{\mathrm{B}=\frac{\mu_0}{4 \pi} \frac{8 \sqrt{2}}{\mathrm{~L}} \mathrm{i}}

\mathrm{\text { So, flux linked with the smaller loop } \phi_2=\mathrm{BS}_2=\left.\frac{\mu_0}{4 \pi} \frac{8 \sqrt{2}}{\mathrm{~L}} \mathrm{i}\right|^2}

\mathrm{\begin{aligned} & \phi_2=\mathrm{Mi}_1 \\ & \text { Or } \quad \mathrm{Mi}=\frac{\mu_0}{4 \pi} \frac{8 \sqrt{2}}{\mathrm{~L}} \mathrm{il}^2 \end{aligned}}

\mathrm{\begin{array}{ll} \text { Or } & M=\frac{\mu_0}{4 \pi} \frac{8 \sqrt{21^2}}{L} \\ \text { Or } & M \alpha 1^2 \end{array}}

Hence, choice (D) is correct and choice (B) is wrong.

 

Posted by

Ritika Kankaria

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