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 :

Question

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 : (A) (2 √2 μ0 L 2/π ℓ) (B) (μ0 ℓ2/2 √2 π L ) (C) (2 √2 μ0 ℓ2/π L ) (D) (μ0 L2/2 √2 π ℓ)

Tardigrade Admin · Accepted Answer

Assuming current I in outer loop magnetic field at centre

= 4 \times \frac{μ _{0} i}{4 π \times \frac{L}{2}} \times (2 sin 4 5^{\circ}) = \frac{2 2 μ _{0} i}{π L}

M = \frac{Flux through inner loop}{i}

M = \frac{2 2 μ _{0} ℓ ^{2}}{π L}

Q. 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 :

$\frac{2 2 μ _{0} L ^{2}}{π ℓ}$

$\frac{μ _{0} ℓ ^{2}}{2 2 π L}$

$\frac{2 2 μ _{0} ℓ ^{2}}{π L}$

$\frac{μ _{0} L ^{2}}{2 2 π ℓ}$

Assuming current I in outer loop magnetic field at centre $= 4 \times \frac{μ _{0} i}{4 π \times \frac{L}{2}} \times (2 sin 4 5^{\circ}) = \frac{2 2 μ _{0} i}{π L}$

$M = \frac{Flux through inner loop}{i}$
$M = \frac{2 2 μ _{0} ℓ ^{2}}{π L}$

Q. 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 :

πℓ22​μ0​L2​

22​πLμ0​ℓ2​

πL22​μ0​ℓ2​

22​πℓμ0​L2​