At any instant, flux passing through the ring is given
\(\text { by } \phi=\mathbf{B} \cdot \mathbf{A}=B A \cos \theta=B A \quad[\because \theta=0]\)
\(\text { or } \phi=B_{0}\left(\pi a^{2}\right) \cos \omega t \quad\left[\because B=B_{0} \cos \omega t\right]\)
By Faraday's law of electromagnetic induction, the magnitude of induced emf is given by

\(e=\frac{d \phi}{d t}=B_{0}\left(\pi a^{2}\right) \omega \sin \omega t\)
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 values of current .at different instants.So, we have current at
\(t=\frac{\pi}{2 \omega} \Rightarrow I=\frac{B_{0}\left(\pi a^{2}\right) \omega}{R} \text { along }\)\(\hat{j}\)
Because, \(\sin \omega t=\sin \left(\omega \frac{\pi}{2 \omega}\right)=\sin \frac{\pi}{2}=1\)
At  \( t=\frac{\pi}{\omega} \Rightarrow I=\frac{B_{0}\left(\pi a^{2}\right) \omega}{R} \times \sin \pi=0 \)
Because \(\sin \omega t=\sin \left(\omega \frac{\pi}{\omega}\right)=\sin \pi=0\)
At \(t=\frac{3 \pi}{2 \omega} \Rightarrow I=\frac{B_{0}\left(\pi a^{2}\right) \omega}{R} \text { along }-\hat{j}\)
\(\sin \omega t=\sin \left(\omega \frac{3 \pi}{2 \omega}\right)=\sin \frac{3 \pi}{2}=-1\)

Here, the hollow cylinder is positively charged.
We know that, the electric field lines appear to come out from the conductor. Thus, the field lines for a uniformly positive charged hollow cylinder is shown in the figure.

The system of magnets will be in stable equilibrium, if the net force on the system is zero and net torque on it is also zero. It will be possible, if the poles of the remaining two magnets is as shown in the figure.

Potential at the centre,
\(V=\frac{1}{4 \pi \Sigma 0}\left(\frac{q}{r}\right)\)
\(=9 \times 10^{9} \times 10^{-8} \times \frac{60}{360} \times 2 \pi r\)
\(=9 \times 10^{+9} \times 10^{-8} \times \frac{2 \times 3.14 \times 2}{6}=188.4 \mathrm{~V}\) 

(i) Electric field between the plates is given by
\(E=\frac{\Delta V}{\Delta x}=-\frac{\left(V_{B}-V_{A}\right)}{1 \times 10^{-2}}\)
\(=\frac{-(0-10)}{10^{-2}}=10^{3} \mathrm{~V} / \mathrm{m}\)
It is directed from A to B
(ii) Work done in moving a charge from A to B,
W_{A\(\rightarrow\)B} = q(\(\triangle\) V) = 20 x 10^-6 (V_B - V_A)
= 20 x 10^-6 (0 -10)
= - 20 x 10^-5 J

(i) As the magnet falls, the magnetic flux linked with the ring increases. This induces emf in the ring which opposes the motion of the falling magnet, hence a < g.
(ii) When current is suddenly switched ON, magnetic flux linked with the solenoid and thus, with metal ring increases. Current is induced in the ring in an anti-clockwise direction (as seen from the top of the ring). Since the direction of flow of current in the ring is opposite to the current in the solenoid, therefore they will repel each other and the ring jumps up.   

Consider the diagram.
Bandwidth = \(\omega _{ 2 }-{ \omega }_{ 1 }\),

Where , \(\omega_1\) and  \(\omega_2\)  correspond to frequencies at which magnitude of  current is \(\frac { 1 }{ \sqrt { 2 } } \) times of maximum value.
\({ I }_{ rms }=\frac { { I }_{ max } }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \approx 0.7A\)
Clearly, from the diagram, the corresponding frequencies are 0.8 rad/s and rad/s.
\(\triangle \omega \) = B and width = 1.2 - 0.8 = 0.4 rad/s

From the first balance point, we get
\(\frac{R}{S}=\frac{33.7}{66.3}\)
After S is connected in parallel with a resistance of 12Ω, the resistance across the gap changes from S to S_eq, where
\(S_{e q}=\frac{12 S}{S+12}\)
and hence the new balance condition now gives
\(\frac{51.9}{48.1}=\frac{R}{S_{e q}}=\frac{R(S+12)}{12 S}\)
Substituting the value of R/S from we get
\(\frac{51.9}{48.1}=\frac{S+12}{12} \cdot \frac{33.7}{66.3}\)
which gives S = 13.5Ω. Using the value of R/S above, we get R = 6.86 Ω.

Here the hollow cylinder is positively charged. We know that the electric field lines appear to come out from the conductor. thus the field lines for a uniformely positive charged hollow cyclinder is shown in the figure.

Here, first we have to find the direction of magnetic field at point O₂ due to the wire carrying current I₁. Use Maxwell's right hand grip (cork screw) rule, the direction of magnetic field at point O₂ due to current I₁ is along Y-axis.
Here,the wire at point O₂ is placed along  Y- axis. Now, by the formula,
F = I₂ (I x B)
Angle between I and B is 0^o ,both are at Y-axis, i.e
 F = IlB sin 0^o = 0
So,the force exerted at point O₂ because of wire along X-axis is zero.