Two concentric circular coils, one of small radius \(r_1\) and the other of large radius \(r_2,\) such that \(r_1\ll r_2\) , are placed co-axially with centers coinciding. Obtain the mutual inductance of the arrangement.
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Two concentric circular coils, one of small radius \(r_1\) and the other of large radius \(r_2,\) such that \(r_1\ll r_2\) , are placed co-axially with centers coinciding. Obtain the mutual inductance of the arrangement.
Answer
Let current \(I_2\) is flowing through larger coil. The magnetic field produced is uniform throughout its area, and the value is given by:
∴ \(B_2=\frac{\mu _0I_2}{2r_2}\)
Since the inner coil is co-axially placed with the outer coil, so the corresponding flux linkage with smaller coil is given by:
\({\emptyset}_1=A_1\times B_2\)
\(=\pi r_1^2\times \frac{\mu _0I_2}{2r_2}\)
\(=\frac{\mu _0\pi r_1^2}{2r_2}I_2\)
\(=M_{12}I_2\)
Where,
\(M_{12}=\frac{\mu _0\pi r_1^2}{2r_2}\); The mutual inductance of the smaller coil with respect to outer coil
Now, we know that,
\(M_{12}=M_{21}=\frac{\mu _0\pi r_1^2}{2r_2}\)
Therefore, the mutual inductance of the arrangement is given by
∴ \(M=M_{12}=M_{21}=\frac{\mu _0\pi r_1^2}{2r_2}\)
∴ \(B_2=\frac{\mu _0I_2}{2r_2}\)
Since the inner coil is co-axially placed with the outer coil, so the corresponding flux linkage with smaller coil is given by:
\({\emptyset}_1=A_1\times B_2\)
\(=\pi r_1^2\times \frac{\mu _0I_2}{2r_2}\)
\(=\frac{\mu _0\pi r_1^2}{2r_2}I_2\)
\(=M_{12}I_2\)
Where,
\(M_{12}=\frac{\mu _0\pi r_1^2}{2r_2}\); The mutual inductance of the smaller coil with respect to outer coil
Now, we know that,
\(M_{12}=M_{21}=\frac{\mu _0\pi r_1^2}{2r_2}\)
Therefore, the mutual inductance of the arrangement is given by
∴ \(M=M_{12}=M_{21}=\frac{\mu _0\pi r_1^2}{2r_2}\)
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