Physics Ncert Solutions Class 12th

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.

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) Initially key K₁ is closed and key K₂ is open, the capacitor C₁ is charged by battery and capacitor C₂ is still uncharged. Now K₁ is opened and K₂ is closed, the capacitors C₁ and C₂ both are connected in parallel. The charge stored by capacitor C₁, gets redistributed between C₁ and C₂ till their potentials become same, i.e., V₂ = V₁.
By law of conservation of charge, the charge stored in capacitor Cx is equal to sum of charges on capacitors C₁ and C₂ when K₁ is opened and K₂ is closed, i.e.,  Q'₁+Q'₂=Q