Physics Ncert Solutions Class 12th

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- in the given figure let C be the amperian loop then,

 

${\int }_Q^{p}H.dl$ = ${\int }_Q^p\frac{B}{m_0}$ .dl

 between B and dl is less than 90 degree so

${\int }_Q^{p}H.dl$ = . ${\int }_Q^p\frac{B}{{\mu }_0}$ dl>0

So magnetic field from south pole to north pole inside the bar magnet

So according to ampere law

${\int }_{pQp}^{}H.dl$ =0

${\int }_{pQp}^{}H.dl$ = $\int H.dl$  +  $\int H.dl$  = 0

${\int }_Q^{p}H.dl$ >0, so ${\int }_p^{Q}H.dl$ <0,

It will be so if the angle between H and dl is more than 90 degree so cos $\theta$  is negative . it means H must run from north to south pole .

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 Questions as classified in NCERT Exemplar

Explanation- I= moment of inertia of the bar magnet

m= mass of bar magnet

l= length of magnet about an any passing through its centre and perpendicular to its

length

M= magnetic moment of the magnet

B= uniform magnetic field in which magnet Is oscillating

T= 2 $\pi \sqrt[]{\frac{I}{MB}}$

I= ml²/12

When magnet is cut into two equal pieces

I¹= $\frac{m{(l/2)}^2}{2\left(12\right)}$ = ml²/96= I/8

Magnetic dipole moment M^'= M/2

Its time period oscillation is T'= 2 $\pi \sqrt[]{\frac{I}{MB}}$ = 2 $\pi \sqrt[]{\frac{I/8}{(M/2)B}}$ = $\frac{2\pi }{2}\sqrt[]{\frac{I}{MB}}$

T'=T/2

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- $\tau =MBsin\theta$

$\tau =pEsin\theta$

From these two we can say that

$MBsin\theta$ = $=pEsin\theta$

pE= MB

E=cB

pcB=MB

p=M/c

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