Electromagnetic Induction Test - 79

Hint: Write total flux through the loop and then use faraday law.

Solution:

Step-1: Find the induced e.m.f

For a coil of $$\mathrm{N}$$ number of turns and area $$\mathrm{A}$$, rotating with the speed $$\omega \mathrm{rad} / \mathrm{sec}$$ in a magnetic field $$\mathrm{B}$$ the induced flux, $$\phi=N B A \sin \omega t$$

$$\therefore$$ Induced e.m.f $$(\varepsilon)=-\frac{d \phi}{d t}=-N B A \omega \cos \omega t$$

Hint: The induced emf in the square frame will be $$\varepsilon  = {B_1}av - {B_2}av$$

Correct Option: (D)

Explanation for Correct Option:

Step 1: Find the magnetic induction.

• As, Magnetic field $$B = \dfrac{{{\mu _0}I}}{{2\pi r}}$$, where $${\mu _0}$$is a constant, $$I$$ is the current through the conductor, and $$r$$is the distance between the conducting points.
• $${B_1} = \dfrac{{{\mu _o}I}}{{2\pi (x - \dfrac{a}{2})}}$$ , where $$I$$ is the current flowing thriugh long wire; $$x$$ is the distance between the long wire and the middle axis of the square frame; $$a$$ is the side length of the square frame.
• $${B_2} = \dfrac{{{\mu _o}I}}{{2\pi (x + \dfrac{a}{2})}}$$

Step 2: Find the induced emf.

Induced emf, $$\varepsilon  = {B_1}av - {B_2}av$$

Here, $$v$$ is the velocity of the frame.

$$ \Rightarrow \varepsilon  = ({B_1} - {B_2})av$$

$$ \Rightarrow \varepsilon  = (\dfrac{1}{{x - \dfrac{a}{2}}} - \dfrac{1}{{x + \dfrac{a}{2}}})\dfrac{{{\mu _o}Iav}}{{2\pi }}$$

$$ \Rightarrow \varepsilon  = \dfrac{{{\mu _o}Iav}}{{2\pi }}\dfrac{a}{{(2x - a)(2x + a)}}$$

Therefore,  the emf induced, $$\varepsilon \alpha \dfrac{1}{{(2x - a)(2x + a)}}$$