Question

Find the induced emf in the square frame as shown in the figure.

• A. $\frac{{\mu }_{0}i}{2\pi }$
• B. $\frac{{\mu }_{0}i}{\pi }\frac{2a^{2}v}{x(x+a)}$
• C. $\frac{{\mu }_{0}i}{\pi }\frac{a^{2}v}{x(x+a)}$
• D. $\frac{{\mu }_{0}i}{2\pi }\frac{a^{2}v}{x(x+a)}$

Answer 1

The correct option is D $\frac{{\mu }_{0}i}{2\pi }\frac{a^{2}v}{x(x+a)}$

The emf induced in the moving conductor is given by,

$E_{induced}=\overrightarrow{l}.(\overrightarrow{v}\times \overrightarrow{B})$

Let $B_1$ be the magnetic field at the side $\text{PS}$.

Let $B_2$ be the magnetic field at the side $\text{RQ}$.

From the diagram, we can see


For the side $\text{PS}$: $l,v$ and $B_1$ are mutually perpendicular.

$\Rightarrow E_{PS}=E_1=lv{B}_1=B_{1}av$

For the side $\text{RQ}$: $l,v$ and $B_2$ are mutually perpendicular.

$E_{RQ}=E_2=B_{2}av$

For the sides, $\text{P}Q$ and $\text{S}R:$ $l$ will be perpendicular to $v\times B$, so the induced emf will be zero.

Now, square frame can be considered as,


$E_{net}=E_1-E_2=av(B_1-B_2)$

The magnetic field due to a straight conductor is given by,

$B=\frac{{\mu }_{0}i}{2\pi x}$

$\therefore B_1=\frac{{\mu }_{0}i}{2\pi x}$ and $B_2=\frac{{\mu }_{0}i}{2\pi (x+a)}$

$\Rightarrow E_{net}=av(\frac{{\mu }_{0}i}{2\pi x}-\frac{{\mu }_{0}i}{2\pi (x+a)})$

$=\frac{av{\mu }_{0}i}{2\pi }\left[\frac{x+a-x}{x(x+a)}\right]=\frac{{\mu }_{0}i}{2\pi }\frac{a^{2}v}{x(x+a)}$

Hence, option $\left(D\right)$ is correct.