When cylindrical bar magnet of radius 0.8 cm is dropped through the metallic pipe with an inner radius of 1 cm , flux linked with the cylinder changes and consequently eddy currents are produced in the metallic pipe. According to Lenz's law, these currents will oppose the (cause) motion of the magnet. Therefore, magnet's downward acceleration will be less than the acceleration due to gravity g. On the other hand, an unmagnetised iron bar will not produce eddy currents and will fall with an acceleration due to gravity $g$. Thus, the magnet will take more time to come down than it takes for a similar unmagnetised cylindrical iron bar dropped through the metallic pipe.

At any instant, flux passes through the ring is given by

$$\phi=\mathrm{B} \cdot \mathrm{~A}=B A \cos \theta=B A\quad (\because\theta=0)$$

or $$\phi=B_0\left(\pi a^2\right) \cos \omega t$$

By Faraday's law of electromagnetic induction.,

Magnitude of induced emf is given by

$$\varepsilon=B_0\left(\pi a^2\right) \omega \sin \omega t$$

This causes flow of induced current, which is given by

$$I=B_0\left(\pi a^2\right) \omega \sin \omega t / R$$

Now, finding the value of current at different instants, so we have current at

$$\begin{aligned} & t=\frac{\pi}{2 \omega} \\ & I=\frac{B_0\left(\pi a^2\right) \omega}{R} \text { along } \hat{\mathrm{j}} \end{aligned}$$

Because

$$\begin{gathered} \sin \omega t=\sin \left(\omega \frac{\pi}{2 \omega}\right)=\sin \frac{\pi}{2}=1 \\ t=\frac{\pi}{\omega} I=\frac{B\left(\pi a^2\right) \omega}{R} \end{gathered}$$

Here, $$\begin{aligned} \sin \omega t & =\sin \left(\omega \frac{\pi}{\omega}\right)=\sin \pi=0 \\ t & =\frac{3}{2} \frac{\pi}{\omega} \\ I & =\frac{B\left(\pi a^2\right) \omega}{R} \text { along }-\hat{\mathbf{j}} \\ \sin \omega t & =\sin \left(\omega \frac{3 \pi}{2 \omega}\right)=\sin \frac{3 \pi}{2}=-1 \end{aligned}$$

The magnetic flux linked with the surface can considered as the number of magnetic field lines passing through the surface. So, let $d \phi=B A$ represents magnetic lines in an area $A$ to B.

By the concept of continuity of lines $B$ cannot end or start in space, therefore the number of lines passing through surface $S_1$ must be the same as the number of lines passing through the surface $S_2$. Therefore, in both the cases we gets the same answer for flux.

The motional electric field $E$ along the dotted line $C D$ ( $\wedge$ to both $\mathbf{v}$ and $\mathbf{B}$ and along $\mathbf{V} \times \mathbf{B}$ ) $=v B$

Therefore, the motional emf along $P Q=($ length $P Q) \times($ field along $P Q)$

$$\begin{aligned} & =(\text { length } P Q) \times(v B \sin \theta) \\ & =\left(\frac{d}{\sin \theta}\right) \times(v B \sin \theta)=v B d \end{aligned}$$

This induced emf make flow of current in closed circuit of resistance $R$. $I=\frac{d v B}{R}$ and is independent of $q$.

The back electromotive force in solenoid is $(u)$ a maximum when there is maximum rate 0 change of current. This occurs is in $A B$ part of the graph. So maximum back emf will be obtained between $5 \mathrm{~s}

Since, the back emf at $t=3 \mathrm{~s}$ is $e$

Also,

the rate of change of current at $t=3, \mathrm{~s}=$ slope of $O A$ from $t=0 \mathrm{~s}$ to $t=5 \mathrm{~s}=1 / 5 \mathrm{~A} / \mathrm{s}$.

So, we have

If $u=L 1 / 5\left(\right.$ for $\left.t=3 \mathrm{~s}, \frac{d I}{d t}=1 / 5\right)$ ( $L$ is a constant). Applying $\varepsilon=-L \frac{d I}{d t}$

Similarly, we have for other values

For $5 \mathrm{~s}

Thus, at $t=7 \mathrm{~s}, u_1=-3 \mathrm{e}$

For $10 \mathrm{~s}

$$u_2=L \frac{2}{20}=\frac{L}{10}=\frac{1}{2} e$$

For $t>30 \mathrm{~s}, u_2=0$

Thus, the back emf at $t=7 \mathrm{~s}, 15 \mathrm{~s}$ and 40 s are $-3 e, e / 2$ and 0 respectively.