Find the current in the sliding rod $A B$ (resistance $=R$ ) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance, $\mathbf{v}$ is constant. Switch $S$ is closed at time $t=0$.
The conductor of length $d$ moves with speed $v$, perpendicular to magnetic field $\mathbf{B}$ as shown in figure. This produces motional emf across two ends of rod, which is given by $=v B d$.
Since, switch $S$ is closed at time $t=0$. current start growing in inductor by the potential difference due to motional emf.
Applying Kirchhoff's voltage rule, we have
$$-L \frac{d I}{d t}+v B d=I R \text { or } L \frac{d I}{d t}+I R=v B d$$
This is the linear differential equation. On solving, we get
$$\begin{aligned} I & =\frac{v B d}{R}+A e^{-R t / 2} \\ \text { At } t & =0 \quad I=0 \\ \Rightarrow\quad A & =-\frac{v B d}{R} \Rightarrow I=\frac{v B d}{R}\left(1-e^{-R t / L}\right) \end{aligned}$$
This is the required expression of current.
A metallic ring of mass $m$ and radius $l$ (ring being horizontal) is falling under gravity in a region having a magnetic field. If $z$ is the vertical direction, the $z$-component of magnetic field is $B_z=B_0(1+\lambda z)$. If $R$ is the resistance of the ring and if the ring falls with a velocity $v$, find the energy lost in the resistance. If the ring has reached a constant velocity, use the conservation of energy to determine $v$ in terms of $m, B$, $\lambda$ and acceleration due to gravity $g$.
The magnetic flux linked with the metallic ring of mass $m$ and radius $l$ falling under gravity in a region having a magnetic field whose $z$-component of magnetic field is $B_z=B_0(1+\lambda z)$ is
$$\phi=B_z\left(\pi l^2\right)=B_0(1+\lambda z)\left(\pi l^2\right)$$
Applying Faraday's law of EMI, we have emf induced given by $\frac{d \phi}{d t}=$ rate of change of flux
Also, by Ohm's law
$$B_0\left(\pi l^2\right) \lambda \frac{d z}{d t}=I R$$
On rearranging the terms, we have $\quad I=\frac{\pi l^2 B_0 \lambda}{R} v$
Energy lost/second $$=I^2 R=\frac{\left(\pi l^2 \lambda\right)^2 B_0^2 v^2}{R}$$
This must come from rate of change in $\mathrm{PE}=m g \frac{d z}{d t}=m g v$ [as kinetic energy is constant for $v=$ constant]
Thus, $$m g v=\frac{\left(\pi l^2 \lambda B_0\right)^2 v^2}{R} \text { or } v=\frac{m g R}{\left(\pi l^2 \lambda B_0\right)^2}$$
This is the required expression of velocity.
A long solenoid $S$ has $n$ turns per meter, with diameter $a$. At the centre of this coil, we place a smaller coil of $N$ turns and diameter $b$ (where $b< a$ ). If the current in the solenoid increases linearly, with time, what is the induced emf appearing in the smaller coil. Plot graph showing nature of variation in emf, if current varies as a function of $m t^2+C$.
Magnetic field due to a solenoid $S, B=\alpha_0 n l$ where signs are as usual.
Magnetic flux in smaller coil $\phi=$ NBA where
$$A=\pi b^2$$
Applying Faraday's law of EMI, we have
$$\begin{aligned} &\text { So, }\\ &\begin{aligned} e & =\frac{-d \phi}{d t}=\frac{-d}{d t}(N B A) \\ & =-N \pi b^2 \frac{d(B)}{d t} \end{aligned} \end{aligned}$$
where,
$$\begin{aligned} B & =\propto_0 N i \\ & =-N \pi b^2 \propto_0 n \frac{d l}{d t} \\ & =-N n \pi \propto_0 b^2 \frac{d}{d t}\left(m t^2+C\right)=-\propto_0 N n \pi b^2 2 m t \end{aligned}$$
$$\begin{aligned} &\text { Since, current varies as a function of } m t^2+C \text {. }\\ &e=-\propto_0 N n \pi b^2 2 m t \end{aligned}$$