Physics - EMI/AC Question with Solution | TestHub

$\text{d}\varphi =\frac{{\mu }_0\text{i}}{2\pi \text{y}}a\text{dy}$

$\varphi =\frac{{\mu }_0\text{i}\text{a}}{2\pi }\underset{\text{x}}{\overset{\text{x}+\text{a}}{\int }}\frac{\text{dy}}{\text{y}}=\frac{{\mu }_0\text{i}\text{a}}{2\pi }\left[\text{ln}\left(\text{x}+\text{a}\right)-\text{ln}\text{x}\right]$

$\text{E.m.f}=-\frac{\text{d}\varphi }{\text{dt}}=-\frac{{\mu }_0\text{i}\mathrm{a}}{2\pi }\left[\frac{1}{\text{x}+\text{a}}\frac{\text{dx}}{\text{dt}}-\frac{1}{\text{x}}\frac{\text{dx}}{\text{dt}}\right]$

$=\frac{{\mu }_{0}ia}{2\pi }\frac{a.v}{x\left(x+a\right)}=\frac{{\mu }_0}{2\pi }.\frac{i{a}^{2}v}{x\left(x+a\right)}=2\times {10}^{-7}\times \frac{1\times \left({0.1}^2\times 10\right)}{0.1\times 0.2}=1\mu V$

Alternative method:

Consider two sides which are perpendicular to the velocity as case of motional e.m.f.


$E=E_1-E{}_2=\left(B_1–B_2\right)V\mathcal{l}$

$=\frac{{\mu }_{0}I}{2\pi }\left(\frac{1}{r_1}-\frac{1}{r_2}\right)V1$

$=2\times {10}^{-7}\times 1(\frac{1}{0.1}-\frac{1}{0.2})10\times 0.1$

$={10}^{-6}v=1\mathrm{ \mu }V$