Electromagnetic Induction

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

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

When the iron core is inserted in the current carrying solenoid, the magnetic field increase due to the magnetisation of iron core and consequently the flux increases.

According to Lenz's law, the emf produced must oppose this increase in flux, which can be done by making decrease in current. So, the current will decrease.

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

When the coil is stretched so that there are gaps between successive elements of the spiral coil i.e., the wires are pulled apart which lead to the flux leak through the gaps.

According to Lenz's law, the emf produced must oppose this decrease, which can be done by an increase in current. So, the current will increase.