Physics

This is a long answer type question as classified in NCERT Exemplar

Since, the magnetic field is brought to zero in time? t, the magnetic flux linked with the ring also reduces from maximum to zero. This, in turn, induces an emf in ring by the

phenomenon of EMI. The induces emf causes the electric field E generation around the

ring.

The induced emf= electric field E (2 πb) (Because V =E d  ) . (i)

By Faraday'slaw of EMI

The induced emf= rate of change of magnetic flux

=rate of change of magnetic field * area= $\frac{B\pi a^2}{dt}$ ……. (ii)

From (i) and (ii)

2 $\pi bE$ =emf= $\frac{B\pi a^2}{dt}$

Since, the charged ring experienced a electric force= QE

This force try to rotate the coil, and the torque is given by

Torque =  b (Force)

= Qeb= Q { $\frac{B\pi a^2}{2\pi bdt}$ }b

= Q $\frac{B\pi a^2}{2dt}$

dL= torque (dt)= Q $\frac{B{a}^2}{2}$

as intital momentum is zero

so final momentum is = mb²w= QBa²/2

w= QBa²/2mb²

This is a long answer type question as classified in NCERT Exemplar

Let us assume that the parallel wires at are y= 0, i.e., along x-axis and y= l. At t= 0, XY has x = 0 i.e., along y-axis.

(i) Let the wire be at x x = (t) at time t.

The magnetic flux linked with the loop is given by

φ = B.A= Bacos0= BA

magnetic flux = B(t)(lx(t))

total emf = emf due to change in field along XYAC+ the motional emf across XY

E= - $\frac{d\varnothing }{dt}$  = - $\frac{dB\left(t\right)}{dt}$ lx(t)- B(t)lv(t)

Electric current in clockwise direction is I= E/r

So force is F= ilBsin90=ilB

Force = $\frac{IB\left(t\right)}{R}$ [- $\frac{dB\left(t\right)}{dt}Ix\left(t\right)-B\left(t\right)Iv\left(t\right)$ ]i

Applying newton 2^nd law

m $\frac{d^{2}x}{d{t}^2}$ = $\frac{I^{2}B\left(t\right)dB}{Rdt}x\left(t\right)$ - $\frac{I^2{B}^2\left(t\right)dx}{Rdt}$

(ii) $\frac{dB}{dt}$ =0

Substituting in eqn 1

$\frac{d^{2}x}{d{t}^2}$ + $\frac{I^2{B}^2\left(t\right)dx}{mRdt}$ =0

$\frac{dv}{dt}+\frac{I^2{B}^{2}v}{mR}$ =o

V= Aexp $(\frac{I^2{B}^{2}v}{mR}$ )=0

(iii) p= I²R= $\frac{I^2{B}^2{v}^{2}R}{R^2}=\frac{B^2{I}^2}{R}{u}_0^2$ exp(-2I²B²tlmR)

This proves decrease in kinetic energy.

This is a long answer type question as classified in NCERT Exemplar

Let us assume that parallel wires at are y=0 i.e along x-axis and y=d.at t=0, AB has x=0 i.e along y-axis and moves with a velocity v

Motional emf across AB is = (B₀sinwt)vd (-j)

Emf due to change in field along OBAC = -B₀wcoswt x (t)d

Total emf in the circuit = emf due to change in field (along OBAC)+ the motional emf across AB= - B₀d {wx coswt+v sinwt}

Electric current in clockwise direction is given by =   $\frac{B_{0}d}{R}$   (wxcoswt+vsinwt)

The force acting on the conductor is given by F= iLbsin90=iLb

Substituting the values

F=   $\frac{B_{0}d}{R}$   (wx coswt+v sinwt) * d * B₀sinwt

  =  $\frac{B_0^2{d}^2}{R}$ (wx coswt+v sinwt)sinwt

This is a multiple choice answer as classified in NCERT Exemplar

(c), (d) Case A: When key K is kept closed and plates of capacitors are moved apart using insulating handle.
The battery maintains the potential difference across connected capacitor in every circumstance. The separation between two plates increases which in turn decreases its capacitance (C=? ₀A/d)and potential difference across connected capacitor continue to be the same as capacitor is still connected with battery. So the charge stored decreases as Q = CV.

Case B: When key K is opened and plates of capacitors are moved apart using insulating handle.The charge stored by isolated charged capacitor remains conserved. The separation between two plates is increasing which in turn decreases its capacitance with the decrease of capacitance, potential difference V increases as V=Q/C.