Electromagnetic Induction MCQs for NEET — Physics Questions with Answers

2. As number of turns are quadrupled, the induced emf will increase four times. Also, resistance of

    coil increases sixteen times. Hence power $\left(\frac{{\mathrm{\epsilon }}^2}{\mathrm{R}}\right)$ will not change.

2. The flux linking with the coil at any instant t is given as $\mathrm{\Phi }=\mathrm{NBA} \cos \mathrm{\omega t}$

$\therefore \mathrm{\epsilon }=-\frac{\mathrm{d\Phi }}{\mathrm{dt}}=\mathrm{NBA\omega sin} \mathrm{\omega t}$

    Therefore, the maximum value of emf is ${\mathrm{\epsilon }}_{\max }=\mathrm{NBA\omega }$

2. Initially, the induced emf (or back emf) is there due to the growth of current through the circuit. The inductance behaves as an open-circuit. Once the current up to its final value ${\mathrm{I}}_0$, there occurs no more change in the current. The back emf induced in the inductance reduces to zero. It behaves like a short-circuit. Hence the final value of the current is ${\mathrm{I}}_0=\frac{\mathrm{E}}{\mathrm{R}} ,$ which depends only on E and R.

2. According to Faraday’s law, an induced emf is set up on the rod whose magnitude is Blv. Thus,

    an electric field is generated in the rod. The electric potential varies uniformly along the rod.

4. The induced emf in the other coil (coil 2) is ${\mathrm{e}}_2=-\mathrm{M}\frac{{\mathrm{di}}_1}{\mathrm{dt}}=-\mathrm{M}\frac{∆{\mathrm{i}}_1}{∆\mathrm{t}} =\mathrm{M}\frac{\left({\mathrm{i}}_2-{\mathrm{i}}_1\right)}{∆\mathrm{t}}=\frac{2\left(0-10\right)}{0.1}=200 \mathrm{V}$

$$\begin{gathered} 4. \mathrm{e}=-\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}} \\ \mathrm{But} \mathrm{e}=4\mathrm{V} \mathrm{and} \frac{\mathrm{di}}{\mathrm{dt}}=\frac{0-1}{{10}^{-3}}=-1/{10}^{-3} \\ \therefore \frac{-1}{{10}^{-3}}\left(-\mathrm{L}\right)=4 \\ \Rightarrow \mathrm{L}=4\times {10}^{-3} \mathrm{henry} \end{gathered}$$

4. Let i be current flowing through the inductance, then flux linked with the circuit $\mathrm{\varphi }\propto \mathrm{i} \mathrm{or} \mathrm{\varphi }=\mathrm{Li}$

$\mathrm{e}=-\frac{\mathrm{d\varphi }}{\mathrm{dt}}=-\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}} \left(\mathrm{emf}\right)$

    Work done against back emf e in time dt and current i is $\mathrm{dW}=-\mathrm{eidt}=\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}} \mathrm{idt}=\mathrm{L} \mathrm{idi}$

$\therefore \mathrm{W}=\mathrm{L} {\int }_0^{\mathrm{i}}\mathrm{i} \mathrm{di}=\frac{1}{2} {\mathrm{Li}}^2$

$$\begin{gathered} 4. \mathrm{\varphi }\propto \frac{1}{{\mathrm{d}}^3} \Rightarrow \frac{{\mathrm{\varphi }}_2}{{\mathrm{\varphi }}_1} ={\left(\frac{{\mathrm{d}}_1}{{\mathrm{d}}_2}\right)}^3 \\ \Rightarrow {\mathrm{\varphi }}_2={\left(\frac{80}{40}\right)}^3\times 4\times {10}^{-5}=32\times {10}^{-5} \mathrm{weber} \\ \therefore ∆\mathrm{\varphi }=28\times {10}^{-5} \mathrm{weber}‐\mathrm{turns} \\ \therefore \mathrm{e}=\frac{∆\mathrm{\varphi }}{∆\mathrm{t}}=\frac{-28\times {10}^{-5}}{8\times {10}^{-2}}=-3.5\times {10}^{-3} \mathrm{V} \\ \mathrm{Thus} \mathrm{emf} \mathrm{produced} =3.5\times {10}^{-3} \mathrm{V} \end{gathered}$$

$$\begin{gathered} 4. \mathrm{Speed} \mathrm{of} \mathrm{train} = \frac{72\times 1000}{3600}=20 \mathrm{m}/\sec \\ \mathrm{Since} \mathrm{unit} \mathrm{of} \mathrm{Earth}\text{'}\mathrm{s} \mathrm{field} \mathrm{is} \mathrm{A}/\mathrm{m}, \mathrm{it} \mathrm{is} \mathrm{magnetic} \mathrm{intensity} \\ \mathrm{Vertical} \mathrm{component} \mathrm{of} \mathrm{earth}\text{'}\mathrm{s} \mathrm{field} \\ \mathrm{Magnetic} \mathrm{intensity} \mathrm{V}=\mathrm{H} \tan \mathrm{\theta }=0.345\times \tan 30° =0.199 \mathrm{A}/\mathrm{m} \\ \mathrm{Magnetic} \mathrm{field} \mathrm{B} = {\mathrm{\mu }}_0\times \mathrm{V} \\ \therefore \mathrm{B}={\mathrm{\mu }}_0\times 0.199 \\ \hspace{0.17em}\mathrm{Hence} \mathrm{induced} \mathrm{emf} \\ \mathrm{e}=4\mathrm{\pi }\times {10}^{-7}\times 0.199\times 1.7\times 20 \\ = 849.8\times {10}^{-6} \mathrm{V} \\ = 850 \mathrm{\mu } \mathrm{V} \end{gathered}$$

$$\begin{gathered} 3. \frac{\mathrm{d\varphi }}{\mathrm{dt}}=10\mathrm{t}+6 \\ \mathrm{e}=-\frac{\mathrm{d\varphi }}{\mathrm{dt}}=-\left(10\mathrm{t}+6\right) \\ \mathrm{e}{|}_{\mathrm{t}=3} =-\left(10\times 3+6\right)=-36 \\ \mathrm{e}{|}_{\mathrm{t}=0} =-\left(10\times 0+6\right)=-6 \\ \frac{{\mathrm{e}}_{\mathrm{t}=3}}{{\mathrm{e}}_{\mathrm{t}=0}} =\frac{-36}{-6} =\frac{6}{1} \end{gathered}$$