$$\eqalign{ & e = - \frac{{\Delta \phi }}{{\Delta t}} = \frac{{ - \Delta \left( {LI} \right)}}{{\Delta t}} = - L\frac{{\Delta I}}{{\Delta t}} \cr & \therefore \left| e \right| = L\frac{{\Delta I}}{{\Delta t}} \Rightarrow 8 = L \times \frac{4}{{0.05}} \cr & \Rightarrow L = \frac{{8 \times 0.05}}{4} = 0.1H \cr} $$

Inductance = $$\frac{{{\mu _0}{N^2}A}}{L}$$

$$\eqalign{ & \phi = 10{t^2} - 50t + 250 \cr & e = - \frac{{d\phi }}{{dt}} = - \left( {20t - 50} \right) \cr & {e_{t = 3}} = - 10V \cr} $$

The current induced will be
$$\eqalign{ & i = \frac{{\left| e \right|}}{R} \Rightarrow i = \frac{1}{R}\frac{{d\phi }}{{dt}}\,\,{\text{But }}i = \frac{{dq}}{{dt}} \cr & \Rightarrow \quad \frac{{dq}}{{dt}} = \frac{1}{R}\frac{{d\phi }}{{dt}} \Rightarrow \int d q = \frac{1}{R}\int d \phi \Rightarrow q = \frac{{BA}}{R} \cr} $$

Velocity of EM wave is given by $$v = \frac{1}{{\sqrt {\mu \in } }}$$
Velocity in air = $$\frac{\omega }{k} = C$$
Velocity in medium = $$\frac{C}{2}$$
Here, $${\mu _1} = {\mu _2} = 1$$   as medium is non-magnetic
$$\therefore \frac{{\frac{1}{{\sqrt {{ \in _{{r_1}}}} }}}}{{\frac{1}{{\sqrt {{ \in _{{r_2}}}} }}}} = \frac{C}{{\left( {\frac{C}{2}} \right)}} = 2 \Rightarrow \frac{{{ \in _{{r_1}}}}}{{{ \in _{{r_2}}}}} = \frac{1}{4}$$

$$\eqalign{ & M = \frac{{{\mu _0}{N_1}{N_2}A}}{\ell } = \frac{{4\pi \times {{10}^{ - 7}} \times 300 \times 400 \times 100 \times {{10}^{ - 4}}}}{{0.2}} \cr & = 2.4\pi \times {10^{ - 4}}H \cr} $$

Emf induced in the coil is given by $$e = - \frac{{d\phi }}{{dt}}$$
If coil has self-inductance $$\left( L \right)$$ and current $$i,$$ then induced emf is given by
$$\eqalign{ & e = - \frac{d}{{dt}}\left( {Li} \right)\,\,{\text{or}}\,\,e = - L\frac{{di}}{{dt}} \cr & \therefore L = \frac{{\left| e \right|}}{{\frac{{di}}{{dt}}}} \cr} $$
$$\eqalign{ & {\text{Given,}}\,\,\left| e \right| = 5\,V,di = 3 - 2 = 1\,A \cr & dt = 1\,ms = 1 \times {10^{ - 3}}s \cr & \therefore L = \frac{{5 \times {{10}^{ - 3}}}}{1} = 5\,mH \cr} $$

KEY CONCEPT: Using $$I = {I_0}\left( {1 - {e^{\frac{{ - t}}{\tau }}}} \right)$$
$$\eqalign{ & {\text{But}}\,{I_0} = \frac{V}{R}\,\,\,{\text{and}}\,\,\,\,\tau = \frac{L}{R} \cr & \therefore I = \frac{V}{R}\left( {1 - {e^{\frac{{ - Rt}}{L}}}} \right) = \frac{{12}}{6}\left[ {1 - {e^{\frac{{ - 6t}}{{8.4}} \times {{10}^{ - 3}}}}} \right] \cr & = 1{\text{ }}\left( {{\text{given}}} \right) \cr & \therefore t = 0.97 \times {10^{ - 3}}s \approx 1ms \cr} $$

KEY CONCEPT :
$$\eqalign{ & P = \frac{{{E^2}}}{R} = \frac{{\pi {r^2}}}{{\rho \ell }}{\left( {\frac{{d\phi }}{{dt}}} \right)^2} = \frac{{\pi {r^2}}}{{\rho \ell }}\left[ {\frac{d}{{dt}}{{\left( {NBA} \right)}^2}} \right] \cr & = \frac{{\pi {r^2}}}{{\rho \ell }}{N^2}{A^2}{\left( {\frac{{dB}}{{dt}}} \right)^2} \Rightarrow P \propto \frac{{{N^2}{r^2}}}{\ell } \cr} $$
Case 1 : $${P_1} \propto \frac{{{N^2}{r^2}}}{\ell },$$     Case 2 : $${P_2} \propto \frac{{{{\left( {4N} \right)}^2}{{\left( {\frac{r}{2}} \right)}^2}}}{{4\ell }}$$
Note : When we decrease the radius of the wire, its length increases but volume remains the same]
$$ \Rightarrow \frac{{{P_1}}}{{{P_2}}} = \frac{1}{1}$$
∴ Power remains the same.

An induced emf (BLv) called motional emf is produced by moving a conductor instead of varying the magnetic field.
For emf, $$e = Bv\left( {{L_{{\text{eff}}}}} \right) = Bv \times \left( {2r} \right) = 2Bvr\,\,\left[ {\because {L_{{\text{eff}}}} = {\text{diameter}} = 2r} \right]$$
$$R$$ will be at higher potential, we can find it by using right hand rule. The electrons of wire will move towards end $$P$$ due to electric force and at end $$R$$ the excess positive change will be left.