A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain.
When the iron core is inserted in the current carrying solenoid, the magnetic field increase due to the magnetisation of iron core and consequently the flux increases. According to Lenz's law, the emf produced must oppose this increase in flux, which can be done by making decrease in current. So, the current will decrease.
Consider a metal ring kept on top of a fixed solenoid (say on a cardboard) (figure). The centre of the ring coincides with the axis of the solenoid. If the current is suddenly switched on, the metal ring jumps up. Explain
When the current is switched on, magnetic flux is linked through the ring. Thus, increase in flux takes place. According to Lenz's law, this increase in flux will be opposed and it can happen if the ring moves away from the solenoid. This happen because the flux increases will cause a counter clockwise current (as seen from the top in the ring in figure.) i.e., opposite direction to that in the solenoid. This makes the same sense of flow of current in the ring (when viewed from the bottom of the ring) and solenoid forming same magnetic pole infront of each other. Hence, they will repel each other and the ring will move upward.
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, $$\begin{aligned} \sin \omega t & =\sin \left(\omega \frac{\pi}{\omega}\right)=\sin \pi=0 \\ t & =\frac{3}{2} \frac{\pi}{\omega} \\ I & =\frac{B\left(\pi a^2\right) \omega}{R} \text { along }-\hat{\mathbf{j}} \\ \sin \omega t & =\sin \left(\omega \frac{3 \pi}{2 \omega}\right)=\sin \frac{3 \pi}{2}=-1 \end{aligned}$$