Physics

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

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.