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  • Find the mutual inductance in the arrangement, when a small circular loop of wire of radius 'R ′ is placed inside a large square loop of wire of side (L >> R). The loops are co

Find the mutual inductance in the arrangement, when a small circular loop of wire of radius 'R ′ is placed inside a large square loop of wire of side (L >> R). The loops are coplanar and their centers coincide :


 

Option: 1

M = \frac{\sqrt{2}\mu_oR^2}{L}


Option: 2

M =\frac{2\sqrt{2}\mu_oR}{L^2}


Option: 3

M =\frac{ \sqrt{2}\mu_oR}{L^2}


Option: 4

M = \frac{2\sqrt{2}\mu_oR^2}{L}


Answers (1)

best_answer

\begin{aligned} & \phi=\mathrm{MI} \\ & \phi_2=\mathrm{MI}_1 \\ & \mathrm{~B}_1 \mathrm{~A}_2=\mathrm{MI}_1 \\ & M=\frac{\mathrm{B}_1 \mathrm{~A}_2}{\mathrm{I}_1} \end{aligned}

\mathrm{B}_1 \rightarrow$ magnetic field due to square frame\\ $\mathrm{A}_2 \rightarrow$ Area of circle\\ $\mathrm{I}_1 \rightarrow$ current in square frame.\\ $\mathrm{B}_1 \rightarrow$\\ $\mathrm{B}_1=4 \cdot \mathrm{B}_{\mathrm{AB}}$\\ $=4\left[\frac{\mu_0 \mathrm{I}_1}{24 \pi \frac{\mathrm{L}}{2}}\left[\sin 45^{\circ}+\sin 45\right]\right]$\\ $\mathrm{B}_1=2 \frac{\mu_0 \mathrm{I}_1}{\pi \mathrm{L}}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=2 \sqrt{2} \frac{\mu_0 I_1}{\pi \mathrm{L}}$\\ $\mathrm{M}=\frac{\mathrm{B}_1 \cdot \mathrm{A}_2}{\mathrm{I}_1}$\\ $\mathrm{M}=\left(\frac{2 \sqrt{2} \mu_0 \mathrm{I}_1}{\pi \mathrm{L}}\right) \times \frac{\pi \mathrm{R}^2}{\mathrm{I}_1}=\frac{2 \sqrt{2} \mu_0 \mathrm{R}^2}{\mathrm{~L}}      

Posted by

Pankaj Sanodiya

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