Physics Ncert Solutions Class 12th

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.

This is a short answer type question as classified in NCERT Exemplar

When the switch is thrown from the off position (open circuit) to the on position (closed circuit), then neither B, nor A nor the angle between B and A change. Thus, no change in magnetic flux linked with coil occur, hence no electromotive force is produced and consequently no current will flow in the circuit.

This is a long answer type question as classified in NCERT Exemplar

This is a long answer type question as classified in NCERT Exemplar

$?=B_z\left(\pi l^2\right)=B_0(1+\lambda z)\left(\pi l^2\right)$

According to faraday's law e= - $\frac{d\varnothing }{dt}$ and also ohm's law V=IR

So by equating these two

We get B_o( $\pi l^2$ ) $\lambda \frac{dz}{dt}=IR$

I= $\frac{\pi l^2{B}_o\lambda v}{R}$

And energy lost/second = I²R= $\frac{{\left(\pi l^2\lambda \right)}^2{B}_o^2{v}^2}{R}$

As change in PE=mg $\frac{dz}{dt}=mgv$ but if we compare both energy then

Mgv= $\frac{{\left(\pi l^2\lambda \right)}^2{B}_o^2{v}^2}{R}$  or v=

This is a long answer type question as classified in NCERT Exemplar

The conductor of length d moves with speed v, perpendicular to magnetic field B as shown in figure. This produces motional emf across two ends of rod, which is given by= vBd.

Since, switch S is closed at time t= 0. current start growing in inductor by the potential

Difference due to motional emf.

Applying Kirchhoff's voltage rule, we have

-L $\frac{dI}{dt}$ +vBd=IR

On solving differential equation we get I= $\frac{vBd}{R}$ + Ae^-Rt/2

At t=0 I=0

A= -vBd/R

I= $\frac{vBd}{R}$ (1-e^-Rt/L)

This is a long answer type question as classified in NCERT Exemplar

The conductor of length d moves with speed v, perpendicular to magnetic field B as shown in figure. This produces motional emf across two ends of rod, which is given by= vB d.

Since, switch S is closed at time t= 0. capacitor is charged by this potential difference. Let

Q ( t) is charge on the capacitor and current flows from A to B.

Now, the induced current I= $\frac{vBd}{R}$ - $\frac{Q}{RC}$

On rearranging  = $\frac{Q}{RC}$ + dQ/dt = $\frac{vBd}{R}$

On solving differential equation we get I= $\frac{vBd}{R}$ e^-t/RC

This is a long answer type question as classified in NCERT Exemplar

The component of magnetic field perpendicular the plane = Bcosθ

But when there is a motional emf then it is e = Blv=vBcosθd

And current Is I= $\frac{\mathrm{v}\mathrm{B}\mathrm{c}\mathrm{o}\mathrm{s}\theta d}{R}$ but when discuss about force F= llB

And component of force which only act is fcosθ

So force is IBd(cosθ) = $\frac{\mathrm{v}\mathrm{B}\mathrm{c}\mathrm{o}\mathrm{s}\theta d}{R}$  where v=dx/dt

Also component of weight is mg cosθ

So by applying newton second law of motion

m $\frac{d^{2}x}{d{t}^2}=mgsin\theta -\frac{Bcos\theta d}{R}\left(\frac{dx}{dt}\right)*\left(Bdcos\theta \right)$

$\frac{dv}{dt}=gsin\theta -\frac{B^2{d}^2}{mR}{\left(cos\theta \right)}^{2}v$

$\frac{dv}{dt}+\frac{B^2{d}^2}{mR}{\left(cos\theta \right)}^{2}v=gsin\theta$

On solving this equation we get

V= $\frac{gsin\theta }{\frac{B^2{d}^2}{mR}{\left(cos\theta \right)}^2}+Aexp(-\frac{B^2{d}^2}{mR}{\left(cos\theta \right)}^{2}t)$