Class 12th

This is a Short Answer Type Question 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 multiple choice answer as classified in NCERT Exemplar

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

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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

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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.