A small square loop of wire of side $l$ is placed inside a large square loop of wire. L(L >> $l$ ). Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is:

Option 1 -

$\frac{2 \sqrt[]{2} μ_{0} L^{2}}{π l}$

Option 2 -

^{$\frac{μ_{0} l^{2}}{2 \sqrt[]{2} π L}$}

Option 3 -

$\frac{2 \sqrt[]{2} μ_{0} l^{2}}{π L}$

Option 4 -

$\frac{μ_{0} L^{2}}{2 \sqrt[]{2 π l}}$

4 Views | Posted 2 months ago

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

Assuming current ‘i’ In outer loop magnetic field at centre

$= 4 \times \frac{μ_{0}}{4 π} \frac{i}{L / 2} \times [2 s i n 45 °)$

$= \frac{2 \sqrt[]{2} μ_{0} i}{π L}$

M = $\frac{F m α t h r o u g h i n n e r l o o p}{i}$ $\frac{}{}$

= $\frac{2 \sqrt[]{2} μ_{0} i l^{2}}{π L} = \frac{2 \sqrt[]{2} μ_{0} l^{2}}{π L}$ $\frac{_{}^{}}{} \frac{_{}^{}}{}$

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A small square loop of wire of side $l$ is placed inside a large square loop of wire. L(L >> $l$ ). Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is:

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A small square loop of wire of side l is placed inside a large square loop of wire. L(L >> l ). Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is:

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A small square loop of wire of side $l$ is placed inside a large square loop of wire. L(L >> $l$ ). Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is: