Magnetism and Matter

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Answer- (c)

Explanation- magnetic field outside the toroid is inversely proportional to distance

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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

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 mag

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 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 ma

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 Questions as classified in NCERT Exemplar

Explanation- the system will be in stable equilibrium if the net force on the system is zero and net torque on the system is also zero.

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

Explanation- magnetic moment of dipole at origin o is

M=Mk

Magnetic field induction at p due to dipole moment

B= $\frac{{\mu }_0}{4\pi }\frac{2Mcos\theta }{r^3}$ r

dS= r(rsin $\theta d\theta$ )r = (r^{2 $\theta$} sind $\theta r$ )

$\int B.ds$  = $\int \frac{{\mu }_0}{4\pi }\frac{2Mcos\theta }{r^3}$ r(r²sin $\theta d\theta r$ )

$\frac{{\mu }_0}{4\pi }\frac{M}{r}{\int }_0^{2\pi }2sin\theta cos\theta d\theta$

$\frac{{\mu }_0}{4\pi }\frac{M}{r}{\int }_0^{2\pi }sin2\theta d\theta$

$\frac{{\mu }_0}{4\pi }\frac{M}{r}\left(\frac{-cos2\theta }{2}\right)$

$\frac{{\mu }_0}{4\pi }\frac{M}{r}$ {cos4 $\pi -cos0$ }

$\frac{{\mu }_0}{4\pi }\frac{M}{r}$ (1-1)= 0

Explanation- When a diamagnetic material is dipped in liquid nitrogen, it again behaves as a diamagnetic material. Thus, superconducting material will again behave as a diamagnetic material. When this diamagnetic material is placed near a bar magnet, it will be feebly magnetised opposite to the dire

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Explanation -  $\mathit{\aleph }$ = I/H

Diamagnetism is due to orbital motion of electrons in an atom developing magnetic moments opposite to applied field. Thus, the resultant magnetic moment of the diamagnetic material is zero and hence, the s