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:
- 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,
$$ \begin{aligned} & M=\frac{\varphi _T}{I}=\frac{\mu _0 a^{2}}{2^{7 / 2} R} \\ \therefore \quad p & =7 \end{aligned} $$
