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A circular wire loop of radius $R$ is placed in the $x-y$ plane centered at the origin $O$. A square loop of side $a(a< < R)$ having two turns is placed with its centre at $z=\sqrt{3} R$ along the axis of the circular wire loop, as shown in figure.
The plane of the square loop makes an angle of $45^{\circ}$ with respect to the $z$-axis. If the mutual inductance between the loops is given by $\frac{\mu_{0} a^{2}}{2^{p / 2} R}$, then the value of $p$ is
The plane of the square loop makes an angle of $45^{\circ}$ with respect to the $z$-axis. If the mutual inductance between the loops is given by $\frac{\mu_{0} a^{2}}{2^{p / 2} R}$, then the value of $p$ is
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The correct answer is:
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The magnetic field due to current carrying wire at the location of square loop is
$B=\frac{\mu_{0}}{4 \pi} \frac{2 \pi i R^{2}}{\left(R^{2}+3 R^{2}\right)^{3 / 2}}=\frac{\mu_{0} i}{16 R}$
The mutual induction
$\begin{array}{l}
M=\frac{N \phi}{i}=\frac{2}{i}\left[\frac{\mu_{0} i}{16 R} \times a^{2} \cos 45^{\circ}\right] \\
\therefore \quad M=\frac{\mu_{0} a^{2}}{2^{\frac{7}{2}} R}
\end{array}$
$B=\frac{\mu_{0}}{4 \pi} \frac{2 \pi i R^{2}}{\left(R^{2}+3 R^{2}\right)^{3 / 2}}=\frac{\mu_{0} i}{16 R}$
The mutual induction
$\begin{array}{l}
M=\frac{N \phi}{i}=\frac{2}{i}\left[\frac{\mu_{0} i}{16 R} \times a^{2} \cos 45^{\circ}\right] \\
\therefore \quad M=\frac{\mu_{0} a^{2}}{2^{\frac{7}{2}} R}
\end{array}$
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