Electromagnetic Induction MCQ Questions & Answers in Electrostatics and Magnetism | Physics

According to Faraday's law of electromagnetic induction, $$\varepsilon = \frac{{d\phi }}{{dt}}$$
$$\eqalign{ & {\text{Also,}}\,\varepsilon = iR \cr & \therefore iR = \frac{{d\phi }}{{dt}} \Rightarrow \int d \phi = R\int i dt \cr} $$
Magnitude of change in flux $$\left( {{\text{d}}\phi } \right) = R \times $$   area under current vs time graph
$${\text{or,}}\,d\phi = 100 \times \frac{1}{2} \times \frac{1}{2} \times 10 = 250Wb$$

Relative velocity $$ = v + v = 2v$$
$$\therefore emf. = B.l\left( {2v} \right)$$

Emf induced in the coil of self-inductance $$\left( L \right)$$ is given by
$$e = - \frac{{d\phi }}{{dt}} = - \frac{d}{{dt}}\left( {Li} \right)\,\,{\text{or}}\,\,e = - L\frac{{di}}{{dt}}\,\,\left( {\frac{{di}}{{dt}} = {\text{rate of flow of current in coil}}} \right)$$
$$\eqalign{ & {\text{As,}}\,\,di = {i_2} - {i_1} = 0 - 10 = - 10\,A \cr & dt = 0.5\,s \cr & e = 220\,V \cr & \therefore 220 = - L\frac{{\left( { - 10} \right)}}{{0.5}} \cr & {\text{or}}\,\,L = \frac{{220}}{{20}} = 11\,H \cr} $$

$$\eqalign{ & \frac{{\Delta \phi }}{{\Delta t}} = \frac{{\left( {{W_2} - {W_1}} \right)}}{t} \cr & {R_{tot}} = \left( {R + 4R} \right)\Omega = 5R\Omega \cr & i = \frac{{nd\phi }}{{{R_{tot}}dt}} = \frac{{ - n\left( {{W_2} - {W_1}} \right)}}{{5Rt}} \cr & \left( {\because {W_2}\,\& \,{W_1}{\text{ are magnetic flux}}} \right) \cr} $$

Let initial e.m.f. induced $$= e.$$
$$\therefore $$ Initial current $$i = \frac{{E - e}}{R}$$
$${\text{i}}{\text{.e}}{\text{.,}}\,2 = \frac{{12 - e}}{1}$$
This gives $$e = 12 - 2 = 10\,{\text{volt}}{\text{.}}$$     As $$e \propto \omega .$$  when speed is halved, the value of induced e.m.f. becomes
$$\frac{e}{2} = \frac{{10}}{2} = 5\,{\text{volt}}$$
$$\therefore $$ New value of current
$$i' = \frac{{E - e}}{R} = \frac{{12 - 5}}{1} = 7\,A$$

According to Faraday's law of electromagnetic induction, $$\varepsilon = \frac{{d\phi }}{{dt}}$$
$$\eqalign{ & {\text{Also,}}\,\,\varepsilon = iR \cr & \therefore iR = \frac{{d\phi }}{{dt}} \cr & \Rightarrow \int {d\phi } = R\int {idt} \cr} $$
Magnitude of change in flux $$\left( {d\phi } \right) = R \times {\text{area}}$$    under current vs time graph
$${\text{or,}}\,d\phi = 100 \times \frac{1}{2} \times \frac{1}{2} \times 10 = 250\,Wb$$

NOTE: Electric field will be induced, as $$ABCD$$  moves, in both $$AD$$ and $$BC.$$  The metallic square loop moves in its own plane with velocity $$v.$$ A uniform magnetic field is imposed perpendicular to the plane of the square loop. $$AD$$ and $$BC$$ are perpendicular to the velocity as well as perpendicular to applied.

WET from $$A$$ to $$B$$ $${\omega _{mg}} = \Delta K.E.$$
$$\eqalign{ & mg\frac{L}{2}\sin \theta = \frac{1}{2}I{\omega ^2} \cr & mg\frac{L}{2}\sin \theta = \frac{1}{2} \times \frac{{m{L^2}}}{3}{\omega ^2} \cr & \frac{{3g}}{L}\sin \theta = {\omega ^2} \cr} $$
Induced emf produced in rod is
$$\eqalign{ & \varepsilon = \frac{1}{2}\left( {B\omega {L^2}} \right) = \frac{1}{2}B \times \sqrt {\frac{{3g\sin \theta }}{L}} \times {L^2} \cr & \varepsilon = \frac{1}{2} \times B \times \sqrt {\frac{{3g\sin \theta \times {L^4}}}{L}} = \frac{1}{2}B\sqrt {3g\sin \theta {L^3}} \cr & \varepsilon \propto {L^{\frac{3}{2}}} \cr & \varepsilon \propto \sqrt {\sin \theta } \cr} $$

Total number of tums, $$N = 20$$
Area of coil, $$A = 25\,c{m^2}$$
$$ = 25 \times {10^{ - 4}}{m^2}$$
Change in magnetic field w.r.t. $$t$$
$$\frac{{dB}}{{dt}} = 1000\,T/s$$
Resistance of coil $$R = 100\,\Omega $$
$$i = ?$$
Induced current, $$i = \frac{e}{R} = \frac{{NA\frac{{dB}}{{dt}}}}{R}\,\,\left[ {e = NA\frac{{dB}}{{dt}}} \right]$$
$$\eqalign{ & = \frac{{20 \times 25 \times {{10}^{ - 4}} \times 1000}}{{100}} \cr & = 0.5\,A \cr} $$

Here, induced e.m.f.

$$e = \int\limits_{2\ell }^{3\ell } {\left( {\omega x} \right)Bdx = B\omega \frac{{\left[ {{{\left( {3\ell } \right)}^2} - {{\left( {2\ell } \right)}^2}} \right]}}{2}} = \frac{{5B{\ell ^2}\omega }}{2}$$