Electromagnetic Induction

Induced emf across the shorter side

$ \begin{aligned} & E_{\text {induced }}=B V l[\because \vec{B} \perp \vec{V} \perp \vec{I}] \\ & =0.3 \times 2 \times 10^{-2} \times 3 \times 10^{-2} \\ & =1.8 \times 10^{-4} \mathrm{~V} \\ & =1.8 \times 10^{-4} \text { volt } \end{aligned} $

Self-inductance of solenoid

$ \begin{aligned} & L=\frac{N \phi}{i} \\ & L=\frac{N B A}{i} \\ & L=\frac{N \mu_0 n i A}{i}\left[n=\frac{N}{I}\right] \\ & L=N \mu_0 n A \\ & L=\left(\frac{N}{I}\right) \mu_0 n A I \\ & L=\mu_0 n^2 A I \\ & L=\mu_0 n^2 \pi r^2 I\left[\because A=\pi r^2\right] \end{aligned} $

$ \begin{aligned} &\text { In figure }(P)\\ &E_P=E_{a b}=-\frac{L d I(t)}{d t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{aligned} &\text { In figure (Q) }\\ &E_Q=E_{a b}=-\frac{L}{2} \frac{d I(t)}{d t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

$ \Rightarrow \frac{E_P}{E_Q}=\frac{2}{1} $

$ \begin{aligned} &\text { } A=A_0(1+\sin t)\\ &\begin{aligned} & \phi=B A=B A_0(1+\sin t) \\ & \varepsilon=-\frac{d \phi}{d t}=-B A_0 \cos t \\ & I_{\text {induced }}=\frac{\varepsilon_{\text {ind }}}{R}=-\frac{B A_0}{R} \cos t \\ & \text { Power }=P=I^2 R=\frac{B^2 A_0^2 \cos ^2 t}{R} \\ & P \propto \cos ^2 t \end{aligned} \end{aligned} $

$\begin{aligned} &\text { Given, } l=2 \mathrm{~A} \text { and } \frac{d i}{d t}=+1 \mathrm{~A} / \mathrm{s}\\ &\begin{aligned} & V_A-L \frac{d i}{d t}-5-i \times 2=V_B \\ & \Rightarrow V_A-1 \times 1-5-2 \times 2=V_B \\ & \Rightarrow V_A-V_B=10 \text { volt } \end{aligned} \end{aligned}$

$\begin{aligned} & L=\mu_0 \mu_r \times n \times A \times N \\ & L=\mu_0 \mu_r n \times A \times \frac{N}{I} \times I \\ & L=\mu_0 \mu_r \times n^2 \times A \times I \Rightarrow L \propto n^2 I \\ & \Rightarrow \frac{L_A}{L_B}=\frac{n_A^2}{n_B^2} \times \frac{I_A}{I_B} \\ & \Rightarrow \frac{L_A}{L_B}=\frac{1}{4} \times 2=\frac{1}{2} \end{aligned}$

$\omega=2\pi \frac{\mathrm{rad}}{\mathrm{sec}}$

$\mathrm{E_{max}=NBA\omega}$

$=100\times0.05\times1\times2\pi$

$=10\times\pi$

$=31.4\,\mathrm{V}$

Magnetic field exist in

Closed Loops (Monopoles do not exist)

$\phi \overrightarrow{\mathrm{B}} \cdot \mathrm{d}\overrightarrow{\mathrm{A}}=0$

(Gauss law for magnetism)

Given magnetic flux is

$\phi = 2{t^3} + 4{t^2} + 2t + 5$

Induced emf is given by

$\varepsilon = {{ - d\phi } \over {dt}} = {{ - d} \over {dt}}(2{t^3} + 4{t^2} + 2t + 5)$

$|\varepsilon | = 6{t^2} + 8t + 2$

At $t = 5$ s,

$|\varepsilon | = 6 \times 25 + 8 \times 5 + 2$

$ = 150 + 40 + 2$

$ = 192$ V

Magnetic flux ($\phi$_B) = $\overrightarrow B $ . $\overrightarrow A $

$\overrightarrow B $ and $\overrightarrow A $ are in same direction, therefore

$\phi$_B = B . A = 0.5 $\times$ 1²

= 0.5 Wb

When a metallic surface is moved in and out in magnetic field, then magnetic flux linked with metallic surface is changed continuously, hence according to Faraday’s law of electromagnetic induction, an induced emf is produced. Since, metallic plate behaves as closed path, therefore an induced current (eddy current) starts flowing in the metallic surface. Hence, assertion and reason both are true and reason is the correct explanation of assertion.

Coil A carries a steady current with increase in temperature, its resistance increases and so current is decreasing at a constant rate, this induces an emf in B which opposes this change i.e., current in coil B is in same direction of A, therefore they attract to each other.

Magnetic flux linked with bigger circular loop is given by

$\phi=\frac{\mu_0}{2} \cdot \frac{\pi I R_1^2 R_2^2}{\left(R_1^2+x^2\right)^{3 / 2}}$

Putting the values,

$\phi=\frac{4 \pi \times 10^{-7} \times \pi \times 15 \times\left(0.3 \times 10^{-2}\right)^2 \times\left(20 \times 10^{-2}\right)^2}{\left[\left(0.3 \times 10^{-2}\right)^2+\left(15 \times 10^{-2}\right)^2\right]^{3 / 2}}$

By solving, we get

$\phi=9.116 \times 10^{-11} \mathrm{~Wb}$