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A long straight wire carrying current $I$ and a rectangular frame with side lengths $a$ and $b$ lie in the same plane as shown in the figure. The mutual inductance of the wire and frame is
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Verified Answer
The correct answer is:
$\frac{\mu_0 b}{2 \pi} \ln 2$
Magnetic field through the wire, $B=\frac{\mu_0 I b}{2 \pi a}$.
$\begin{array}{ll}\therefore \text { Magnetic flux induced, } \phi & =\oint_s \mathbf{B} \cdot d \mathbf{a} \\ \Rightarrow & \phi=\frac{\mu_0 I b}{2 \pi} \int_a^{a+a} \frac{d a}{a} \\ \Rightarrow & \phi=\frac{\mu_0 I b}{2 \pi} \ln 2\end{array}$
$\therefore$ Mutual inductance of the wire and the frame is
$$
M=\frac{\phi}{I}=\frac{\mu_0 b}{2 \pi} \ln 2
$$
$\begin{array}{ll}\therefore \text { Magnetic flux induced, } \phi & =\oint_s \mathbf{B} \cdot d \mathbf{a} \\ \Rightarrow & \phi=\frac{\mu_0 I b}{2 \pi} \int_a^{a+a} \frac{d a}{a} \\ \Rightarrow & \phi=\frac{\mu_0 I b}{2 \pi} \ln 2\end{array}$
$\therefore$ Mutual inductance of the wire and the frame is
$$
M=\frac{\phi}{I}=\frac{\mu_0 b}{2 \pi} \ln 2
$$
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