Consider a metal ring kept (supported by a cardboard) on top of a fixed solenoid carrying a current $I$ (see figure of Question 14). The centre of the ring coincides with the axis of the solenoid. If the current in the solenoid is switched off, what will happen to the ring?
When the current is switched off, magnetic flux linked through the ring decreases. According to Lenz's law, this decrease in flux will be opposed and the ring experience downward force towards the solenoid. This happen because the flux $i$ decrease will cause a clockwise current (as seen from the top in the ring in figure) i.e., the same direction to that in the solenoid. This makes the opposite sense of flow of current in the ring (when viewed from the bottom of the ring) and solenoid forming opposite magnetic pole infront of each other. Hence, they will attract each other but as ring is placed at the cardboard it could not be able to move downward.
Consider a metallic pipe with an inner radius of 1 cm . If a cylindrical bar magnet of radius 0.8 cm is dropped through the pipe, it takes more time to come down than it takes for a similar unmagnetised cylindrical iron bar dropped through the metallic pipe. Explain.
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.
A magnetic field in a certain region is given by $\mathbf{B}=B_0 \cos (\omega t) \hat{\mathbf{k}}$ and $a$ coil of radius a with resistance $R$ is placed in the $x-y$ plane with its centre at the origin in the magnetic field (figure). Find the magnitude and the direction of the current at $(a, 0,0)$ at
$$t=\frac{\pi}{2 \omega}, t=\frac{\pi}{\omega} \text { and } t=\frac{3 \pi}{2 \omega}$$
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, $$\begin{aligned} \sin \omega t & =\sin \left(\omega \frac{\pi}{\omega}\right)=\sin \pi=0 \\ t & =\frac{3}{2} \frac{\pi}{\omega} \\ I & =\frac{B\left(\pi a^2\right) \omega}{R} \text { along }-\hat{\mathbf{j}} \\ \sin \omega t & =\sin \left(\omega \frac{3 \pi}{2 \omega}\right)=\sin \frac{3 \pi}{2}=-1 \end{aligned}$$
Consider a closed loop $C$ in a magnetic field (figure). The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula $\phi=\mathbf{B}_1 d \mathbf{A}_1, \mathbf{B}_2 d \mathbf{A}_2 \ldots .$. . Now, if we choose two different surfaces $S_1$ and $S_2$ having $C$ as their edge, would we get the same answer for flux. Justify your answer.
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.
Find the current in the wire for the configuration shown in figure. Wire $P Q$ has negligible resistance. $B$, the magnetic field is coming out of the paper. $\theta$ is a fixed angle made by $P Q$ travelling smoothly over two conducting parallel wires separated by a distance $d$.
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 $q$.