Physics NCERT Exemplar Solutions Class 12th Chapter Six

This is a multiple choice answer as classified in NCERT Exemplar

(a) When the current in B (at t= 0) is counter-clockwise and the coil A is considered above to it. The counterclockwise flow of the current in B is equivalent to north pole of magnet and magnetic field lines are emanating upward to coil A. When coil A start rotating at (t= 0), the current in A is constant along clockwise direction by Lenz's rule.

This is a multiple choice answer as classified in NCERT Exemplar

(b) When the A stops moving the current in B become zero, it is possible only if the current in A is constant. If the current in A would be variable, there must be an induced emf (current) in B even if the A stops moving.

This is a multiple choice answer as classified in NCERT Exemplar

(b) When cylindrical bar magnet is rotated about its axis, no change in flux linked with the circuit takes place, consequently no emf induces and hence, no current flows through the ammeter A.

This is a multiple choice answer as classified in NCERT Exemplar

$\varnothing =B.A$

And A= A₁+A₂= (L²k+L²i)

B= B_o (i+k)T

$\varnothing =B.A$= B_o (i+k) (L²k+L²i)

= 2B_oL²Wb

This is a multiple choice answer as classified in NCERT Exemplar

(c) $\varnothing$ =B.A= B₀ (2i+3j+4k)= 4B₀L²wb

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

Mutual inductance of coil A with respect to B

M₂₁=N₂ $\varnothing$ ₂/I₁ = $\frac{{10}^{-2}}{2}$ = 5mH

N₁ $\varnothing$ ₁= M₁₂I₂= 5mH (1A)= 5mWb

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

The back emf in solenoid force in solenoid is U a maximum rate of change of current . so maximum back emf will be obtained between 5s

Since the back emf at t = 3s also the rate of change of current at t= 3s, s= slope of OA from t=0s to t= 5s=1/5 A/sec

So we have if u= L1/5 (for t= 3s, dI/dt=1/5) (L is a constant). Applying e=-LdI/dt

For 5s

At t= 7s, u₁=-3e

For 10s

For t>30s, u₂=0

Thus back emf at t=7s,15s and 40s are -3e, e/2 and 0 respectively.

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

The motional emf along PQ = length PQ * field along PQ

= length PQ *  vBsin $\mathrm{\theta }$

= $\frac{d}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{\theta }\mathrm{}}*$ vB sin $\mathrm{\theta }$ = vBd

So emf make the flow of current in the circuit with resistance R

I= dvB/R

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

The magnetic flux linked with the surface can considered as the number of magnetic field lines passing through the surface. So, let dφ = BA represents magnetic lines in an area to. By the concept of continuity of lines cannot end or start in space, therefore the number of lines passing through surface S₁ must be the same as the number of lines passing through the surface S₂.Therefore, in both the cases we gets the same answer for flux.

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

 $\varnothing =B.A=BAcos\theta$

$\varnothing =B_o\pi a^{2}coswt$

But according to faraday's law e= B_{o $\pi a^{2}wsinwt$}

So the current will be I=B_{0 $\frac{\pi a^{2}wsinwt}{R}$}

For t= $\frac{\pi }{2w}$

I= $\frac{B_o\pi a^{2}w}{R}$ along j

Sinwt= sin(w $\frac{\pi }{2w}$ ) = sin $\frac{\pi }{2}$ =1

T= $\frac{\pi }{w}$ , I= $\frac{B\left(\pi a^2\right)w}{R}$

Sinwt=sinw $\frac{\pi }{w}$  = sin $\pi =0$

T= $\frac{3\pi }{2w}$

I= $\frac{B_o\pi a^{2}w}{R}$

So it become sin $\frac{3\pi }{2w}$ =-1