Electromagnetic Induction

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

6.16 Let us take a small element dy in the loop, at a distance y from the long straight wire.

Magnetic flux associated with element dy, d $\varnothing$ = BdA, where

dA = Area of the element dy = a dy

Magnetic flux at distance y, B = $\frac{{\mu }_{0}I}{2\pi y}$ , where

I = current in the wire

${\mu }_0$ = Permeability of free space = 4 $\pi *{10}^{-7}$ T m $A^{-1}$

Therefore,

d $\varnothing$ = $\frac{{\mu }_{0}I}{2\pi y}$ a dy = $\frac{{\mu }_{0}Ia}{2\pi }$ $\frac{dy}{y}$

$\varnothing =\frac{{\mu }_{0}Ia}{2\pi }\int \frac{dy}{y}$

Now from the figure, the range of y is x to x+a. Hence,

$\varnothing =\frac{{\mu }_{0}Ia}{2\pi }{\int }_x^{x+a}\frac{dy}{y}$ = $\frac{{\mu }_{0}Ia}{2\pi }{\left[{\log }_e?y\right]}_x^{a+x}$ = $\frac{{\mu }_{0}Ia}{2\pi }{\log }_e?(\frac{a+x}{x})$

For mutual inductance M, the flux is given as

$\varnothing =MI$ . Hence

$MI=\frac{{\mu }_{0}Ia}{2\pi }{\log }_e?(\frac{a+x}{x})$

M = $\frac{{\mu }_{0}a}{2\pi }{\log }_e?(\frac{a}{x}+1)$

Emf induced in the loop, e = Bav

= ( $\frac{{\mu }_{0}I}{2\pi x}$ ) $*$ av

For I = 50 A, x = 0.2 m, a = 0.1 m, v = 10 m/s

e = $\frac{4\pi *{10}^{-7}*50*0.1*10}{2\pi *0.2}$ = 5 $*{10}^{-5}$ V

6.14 Length of the rod, l = 15 cm = 0.15 m

Magnetic field strength, B = 0.5 T

Resistance of the closed loop, R = 9 mΩ = 9 $*{10}^{-3}$ Ω

Induced emf = 9 mV

Polarity of the induced emf is such that end P shows positive while end Q shows negative ends.

Speed of the rod, v = 12 cm/s = 0.12 m/s

Induced emf is given as, e = Bvl = 0.5 $*0.12*0.15$ = 9 mV

Yes, when key K is closed, excess charge is maintained by the continuous flow of current. When the key K is open, there is excess charge built up at both ends of the rods.

Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod. There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.

Retarding force exerted on the rod, F = IBl , where

I = Current flowing through the rod = $\frac{e}{R}$ = $\frac{9*{10}^{-3}}{9\mathrm{}*{10}^{-3}}$ = 1 A

F = 1 $*0.5*0.15$ = 0.075 N

Speed of the rod, v = 12 cm/s = 0.12 m/s

Power is given by P = Fv = 0.075 $*0.12=9*{10}^{-3}$ W = 9 mW

When key K is open, no power is expended.

Power dissipated by heat = $I^{2}R=1*$ 9 $*{10}^{-3}$ W = 9 mW

The source of this power is an external agent.

In this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.

6.13 Area of the coil, A = 2 ${cm}^2$ = 2 $*{10}^{-4}$ $m^2$

Number of turns, n = 25

Total charge flowing in the coil, Q = 7.5 mC = 7.5 $*{10}^{-3}$ C

Total resistance of the coil and galvanometer, R = 0.50 Ω

We know, Induced current in the coil, I = $\frac{Inducedemf\left(e\right)}{R}$ ……… (1)

Induced emf is given by the equation, e = $-$ n $\frac{d\varnothing }{dt}$ ………………… (2),

Where d $\varnothing$ is the change of flux

Combining equations (1) and (2), we get

I = $\frac{-\mathrm{n}\frac{d\varnothing }{dt}}{R}$ or Idt = $-\frac{n}{R}$ d $\varnothing$ ……… (3)

Initial flux through coil, ${\varnothing }_i$ = BA, where B = Magnetic field strength

Final flux through coil, ${\varnothing }_f$ = 0

Integrating equation (3), we get

$\int Idt$ = $-\frac{n}{R}{\int }_{{\varnothing }_i}^{{\varnothing }_f}d\varnothing$

$\int Idt=Q=\mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{}\mathrm{f}\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{i}\mathrm{l}$

Therefore

Q = $-\frac{n}{R}\left({\varnothing }_f-{\varnothing }_i\right)=-\frac{n}{R}\left(0-{\varnothing }_i\right)=\frac{n{\varnothing }_i}{R}=\frac{nBA}{R}$

B = $\frac{QR}{nA}$ = $\frac{7.5\mathrm{}*{10}^{-3}*0.5}{25*2*{10}^{-4}}$ = 0.75 T

Hence, the field strength is 0.75 T

6.12 Side of the square loop, s = 12 cm = 0.12 m

Area of the square loop, A = 0.12 $*0.12=0.0144m^2$

Velocity of the loop, v = 8 cm /s = 0.08 m/s

Gradient of the magnetic field along negative x – direction.

$\frac{dB}{dx}$ = ${10}^{-3}$ T/cm = ${10}^{-1}$ T/m

Rate of decrease of magnetic field

$\frac{dB}{dt}$ = ${10}^{-3}$ T/s

Resistance of the loop, R = 4.50 mΩ = 4.5 $*{10}^{-3}$ Ω

Rate of change of the magnetic flux due to motion of the loop in a non-uniform magnetic field is given as:

$\frac{d\varnothing }{dt}$ = A $*\frac{dB}{dx}$ $*v$ = $0.0144*{10}^{-1}*0.08$ = 1.152 $*{10}^{-4}$ T $m^2{s}^{-1}$

Rate of change of flux due to explicit time variation in field B is given as:

$\frac{d\varnothing \text{'}}{dt}$ = A $*\frac{dB}{dt}$ = $0.0144*{10}^{-3}$ = 1.44 $*{10}^{-5}$ T $m^2{s}^{-1}$

Since the rate of change of the flux is the induced emf, the total emf in the loop can be calculated as:

e = 1.152 $*{10}^{-4}$ + 1.44 $*{10}^{-5}$ = 1.296 $*{10}^{-4}$ V

The induced current, I = $\frac{e}{R}$ = $\frac{1.296\mathrm{}*{10}^{-4}}{4.5\mathrm{}*{10}^{-3}}$ = 28.8 $*{10}^{-3}$ A

Therefore, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.

6.11 The area of the rectangular coil, A = 8 $*2=$ 16 ${cm}^2$ = 16 $*{10}^{-4}$ $m^2$

Initial value of the magnetic field, $B_1$ = 0.3 T

Rate of decrease of the magnetic field, $\frac{dB}{dt}$ = 0.02 T/s

From the relation of induced emf e = $\frac{d\varnothing }{dt}$ , where $d\varnothing$ is the change in the flux linkage with the coil = A $*B$

Hence, e = $\frac{d\left(AB\right)}{dt}$ = A $\frac{dB}{dt}$ = 16 $*{10}^{-4}*0.02$ = 3.2 $*{10}^{-5}$ V

Resistance in the loop, R = 1.6 Ω

Hence I = $\frac{e}{R}$ = $\frac{3.2\mathrm{}*{10}^{-5}}{1.6}$ = 2 $*{10}^{-5}$ A

Power dissipated in the form of heat is given by

P = $I^2$ R = ( ${\mathrm{}2\mathrm{}*{10}^{-5})}^2*1.6$ = 6.4 $*{10}^{-10}$ W