Explanation
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 $$(\varepsilon)=-\frac{d \phi}{d t}=-N B A \omega \cos \omega t$$
Step-2: Find the peak current
Induced e.m.f $$(\varepsilon)=-N B A \omega \cos \omega t$$
$$\therefore$$ Peak value of e. $$m \cdot f=N B A \omega$$
Therefore peak current through the resistance
$$=\frac{\text { Peak e.m.f }}{\text { Total resistance }}=\frac{N B A \omega}{5+10}$$ $$\left(\begin{array}{l}\text { Given, } \mathrm{N}=500 \\ \mathrm{~B}=0.14 \mathrm{~Wb} / \mathrm{m}^{2} \\ A=50 \mathrm{~cm}^{2}=50 \times 10^{-4} \mathrm{~m}^{2} \\ \omega=150 \mathrm{rad} / \mathrm{sec}\end{array}\right)$$
$$=\frac{500 \times 0.14 \times\left(50 \times 10^{-4}\right) \times 150}{15}$$
$$=3.5 \mathrm{~A}$$
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.
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, $$\varepsilon \alpha \dfrac{1}{{(2x - a)(2x + a)}}$$
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