Class 12th

Kindly go through the solution

Kindly go through the solution

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 .

Kindly go through the solution

Kindly go through the solution

This is a Short Answer Type Question 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 $\pi \sqrt[]{\frac{I/8}{(M/2)B}}$ = $\frac{2\pi }{2}\sqrt[]{\frac{I}{MB}}$

Kindly go through the solution

Kindly go through the solution

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