Question

(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig.
(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.

Answer 1

Take a small element $dy$ in the loop at a distance $y$ from the long straight wire.

Magnetic flux associated with element, $dy$ is $d\varphi =BdA$

Where, $dA=a\times dy$ is Area of element $dy$

$B={\mu }_{o}I/2\pi y$ is magnetic field at $y$.

$I$ is current in the wire.

$\therefore d\varphi ={\mu }_{o}Ia\times dy/2\pi y$

$\varphi =\left({\mu }_{o}Ia/2\pi \right){\int }_x^{a+x}dy/y$

$\Rightarrow \varphi =\left({\mu }_{o}Ia/2\pi \right)ln\left(\frac{a+x}{x}\right)$

Mutual Inductance $M=\varphi /I=\left({\mu }_{o}a/2\pi \right)ln\left(\frac{a+x}{x}\right)$

Emf induced in the loop, $e=Bav=\left({\mu }_{o}I/2\pi x\right)av$

From the given values, $e=5\times {10}^{-5}V$