Electromagnetic Induction MCQs for NEET — Physics Questions with Answers

$$\begin{gathered} 1. \mathrm{\varphi }=\mathrm{Li} \Rightarrow \mathrm{NBA}=\mathrm{Li} \\ \mathrm{Since} \mathrm{magnetic} \mathrm{field} \mathrm{at} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{circular} \mathrm{coil} \mathrm{carrying} \mathrm{current} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{2\mathrm{\pi Ni}}{\mathrm{r}} \\ \therefore \mathrm{N}.\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}. \frac{2\mathrm{\pi Ni}}{\mathrm{r}}. {\mathrm{\pi r}}^2=\mathrm{Li} \Rightarrow \mathrm{L}=\frac{{\mathrm{\mu }}_0{\mathrm{N}}^2\mathrm{\pi r}}{2} \\ \mathrm{Hence} \mathrm{self} \mathrm{inductance} \mathrm{of} \mathrm{a} \mathrm{coil} \frac{4\mathrm{\pi }\times {10}^{-7}\times 500\times 500\times \mathrm{\pi }\times 0.05 }{2} =25 \mathrm{mH} \end{gathered}$$

1.  Magnetic induction in the solenoid

     $\mathrm{B} = {\mu }_0\mathrm{ni}$

     Magnetic flux linked with the coil

     $\mathrm{\varphi } = \mathrm{NAB} = \mathrm{NA}{\mu }_0\mathrm{ni}$

$$\begin{gathered} \mathrm{M}=\frac{\mathrm{\varphi }}{\mathrm{i}}=\frac{\mathrm{N} \mathrm{A} {\mathrm{\mu }}_0\mathrm{n} \mathrm{i}}{\mathrm{i}}={\mathrm{NA\mu }}_0\mathrm{n} \\ \mathrm{M}=100\times \mathrm{\pi }(1 \times {10}^{-2}{)}^2\times 4\mathrm{\pi }\times {10}^{-7}\times 800= 316\times {10}^{-7} \mathrm{H}=0.0316 \mathrm{mH}. \end{gathered}$$

$$\begin{gathered} 3. \mathrm{The} \mathrm{magnetic} \mathrm{field} \mathrm{at} \mathrm{a} \mathrm{distance} \mathrm{x} \mathrm{from} \mathrm{the} \mathrm{wire} \mathrm{B}\left(\mathrm{x}\right)=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi x}} \\ \mathrm{EMF} \mathrm{induced} \mathrm{in} \mathrm{an} \mathrm{element} \mathrm{of} \mathrm{length} \mathrm{dx} \mathrm{at} \mathrm{a} \mathrm{distance} \mathrm{x} \mathrm{from} \mathrm{wire}=\mathrm{Bvdx} \\ \therefore \mathrm{Total} \mathrm{EMF} \mathrm{induced} \mathrm{in} \mathrm{the} \mathrm{rod} \\ \mathrm{E}=\underset{0.01}{\overset{0.2}{\int }}\mathrm{Bv} \\ \mathrm{dx}=\underset{0.01}{\overset{0.2}{\int }}\frac{{\mathrm{\mu }}_0\mathrm{iv}}{2\mathrm{\pi x}} \\ \mathrm{dx}=\frac{{\mathrm{\mu }}_0\mathrm{iv}}{2\mathrm{\pi }} \underset{0.01}{\overset{0.2}{\int }}\frac{1}{\mathrm{x}} \mathrm{dx} \\ \mathrm{E}=\frac{{\mathrm{\mu }}_0\mathrm{iv}}{2\mathrm{\pi }}{\left[{\log }_{\mathrm{e}} \mathrm{x}\right]}_{0.01}^{0.2} \\ = \frac{{\mathrm{\mu }}_0\mathrm{iv}}{2\mathrm{\pi }}\left[{\log }_{10}\left(0.2\right)-{\log }_{10}\left(0.01\right)\right]\times 2.303 \\ \mathrm{E}=\frac{4\mathrm{\pi }\times {10}^{-7}\times 5\times 10}{2\mathrm{\pi }}\left[1.301\right]\times 2.303 \\ =2.99\times {10}^{-5} \mathrm{V}\approx 30\mathrm{\mu V} \end{gathered}$$

 $$\begin{gathered} 4. \mathrm{Flux} \mathrm{linked} \mathrm{with} \mathrm{the} \mathrm{search} \mathrm{coil} \mathrm{\varphi }={\mathrm{BAN}}_{\mathrm{s}}={\mathrm{\mu }}_0{\mathrm{niAN}}_{\mathrm{s}} \\ \therefore \frac{\mathrm{d\varphi }}{\mathrm{dt}}={\mathrm{\mu }}_0{\mathrm{nAN}}_{\mathrm{s}}\frac{\mathrm{di}}{\mathrm{dt}} \\ \mathrm{i}={\mathrm{i}}_0 \sin \mathrm{\omega t} \\ \therefore \frac{\mathrm{d\varphi }}{\mathrm{dt}}={\mathrm{\mu }}_0{\mathrm{nAN}}_{\mathrm{s}}{\mathrm{i}}_0\mathrm{\omega } \cos \mathrm{\omega t} \\ {\mathrm{E}}_{\max }={\left(\frac{\mathrm{d\varphi }}{\mathrm{dt}}\right)}_{\max }={\mathrm{\mu }}_0{\mathrm{nAN}}_{\mathrm{s}}{\mathrm{i}}_0\mathrm{\omega } \\ \therefore \mathrm{f}=\frac{\mathrm{\omega }}{2\mathrm{\pi }}=\frac{{\mathrm{E}}_{\max }}{2{\mathrm{\pi \mu }}_0{\mathrm{nAN}}_{\mathrm{s}}{\mathrm{i}}_0} \\ \mathrm{f}=\frac{2.5\times {10}^{-2}}{6.28\times 12.56\times {10}^{-7}\times {10}^5\times {10}^{-4}\times 20\times 1} =15.85 {\mathrm{s}}^{-1} \end{gathered}$$

$$\begin{gathered} 1. \mathrm{E}=-\mathrm{NA}\frac{\mathrm{dB}}{\mathrm{dt}} \\ \therefore \mathrm{i}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{\mathrm{NA}}{\mathrm{R}}\frac{\mathrm{d}}{\mathrm{dt}}\left(75{\mathrm{e}}^{-200\mathrm{t}}\right)\times {10}^{-4} \\ =\frac{\mathrm{NA}}{\mathrm{R}}\left[75\times \left(-200\right){\mathrm{e}}^{-200\mathrm{t}}\right]\times {10}^{-4} \\ =+\frac{50\times 7\times {10}^{-8}}{30}\left[15000{\mathrm{e}}^{-1}\right] \\ =\frac{175\times {10}^{-5}}{\mathrm{e}}=\frac{175\times {10}^{-5}}{2.73}=0.64\times {10}^{-3} \mathrm{A} \\ \mathrm{i}=0.64 \mathrm{mA} \end{gathered}$$

According to the formula of Magnetic flux 
Magnetic flux $\varphi$=B.A and A=$\pi r^2$ 
It can be also written as = B. $\pi r^2$ 
Therefore Induced emf can be found by using the relation magnetic flux as follows- 
$\left|e\right|=\frac{d\varphi }{dt}=B\left(2\pi r\right)\frac{dr}{dt}$ 
Now we put the given value in the equation 
= 0.025 x $\pi$x 2 x 2 10^-2 x 1 x 10^-3 = $\pi \mu V$

Given, 
Self-inductance of coil 1 = 2 mH 
Self-inductance of coil 2 = 8 mH 
When the total flux associated with one coil links with the other i.e., a case of maximum flux linkage, then 
M₁₂ =$\frac{N_2{\varphi }_{B_2}}{i_1}$ and M₂₁ =$\frac{N_1\varphi B_1}{i_2}$ 
Similarly, L₁ =$\frac{N_1{\varphi }_{B_1}}{i_1}$ and L₂=$\frac{N_2{\varphi }_{B_2}}{i_2}$ 
If all the flux of coil 2 links coil 1 and vice-versa then 
${\varphi }_{B_2}$ = ${\varphi }_{B_1}$
Since, M₁₂ = M₂₁ = M, hence we have 
M₁₂M₂₁=M² =$\frac{N_1{N}_2{\varphi }_{B_1}{\varphi }_{B_2}}{i_1{i}_2}$ = L₁L₂ 
∴ M_max =$\sqrt{L_1{L}_2}$ 
Given, L₁ = 2 mH, L₂ = 8 mH 
∴ M_max = $\sqrt{2\times 8}=\sqrt{16}$= 4 mH