Electromagnetic Induction

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

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

applying ohm's law

I=E/R=d $\varnothing /Rdt$

I= dQ/dt or dQ/dt= d $\varnothing /Rdt$

Q (t₁)-Q (t₂)= $\frac{1}{R}$ [ $\varnothing$ t₁- $\varnothing$ t₂]

$\varnothing$ t₁= L_{1 $\frac{{\mu }_0}{2\pi }{\int }_x^{L_2+x}\frac{dx}{x}I$} t₁

= $\frac{{\mu }_0{L}_1}{2\pi }$ I1in [ $\frac{L_2+x}{x}$ ]

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

B= $\frac{{?}_{0}I}{2R}$ (ouside the paper)

Flux= $\frac{{?}_{0}I}{2R}l{?}_{x_0}^x\frac{dr}{r}$ = $\frac{{?}_{0}I}{2R}$ in $\frac{x}{x_0}$

Induced emf= e/R=I

Induced current = $\frac{1}{R}\frac{d?}{dt}$ = $\frac{e}{R}$ = $\frac{{?}_{0}I?}{2?R}$ in $\frac{x}{x_0}$

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

At t=o t=T/8= $\pi /4w$  the rod will make contact with side BD.at any time 0 $\pi /4w$

So magnetic flux through area ODQ is $\varnothing =B*areaOQD=B*\frac{1}{2QD}*OD$

=B $*\frac{1ltan\theta }{2}*l$

$\varnothing =\frac{1}{2}B{L}^{2}wtan\theta =\frac{1}{2}B{L}^{2}w{sec}^{2}wt$

I=e/R

And R= $\lambda x=\frac{\lambda l}{coswt}$  so I= $\frac{1}{2}\frac{BLw}{coswt\lambda }$

Same result for other side too