Electromagnetic Induction and Alternating Current 3 Question 8

####9. A circular wire loop of radius $R$ is placed in the $x-y$ plane centred at the origin $O$. A square loop of side $a(a \ll R)$ 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

(2012)

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Answer:

Correct Answer: 9. 7

Solution:

  1. If $I$ current flows through the circular loop, then magnetic flux at the location of square loop is

$$ B=\frac{\mu_{0} I R^{2}}{2\left(R^{2}+Z^{2}\right)^{3 / 2}} $$

Substituting the value of $Z(=\sqrt{3} R)$,

we have

$$ B=\frac{\mu_{0} I}{16 R} $$

Now, total flux through the square loop is

$$ \varphi_{T}=N B S \cos \theta=(2) \frac{\mu_{0} T}{16 R} \quad a^{2} \cos 45^{\circ} $$

Mutual inductance,

$$ \begin{aligned} & M=\frac{\varphi_{T}}{I}=\frac{\mu_{0} a^{2}}{2^{7 / 2} R} \\ \therefore \quad p & =7 \end{aligned} $$



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