Practice questions for electromagnetic Induction

During the motion of the Jumper , The magnetic Flux across the loop closed by the Jumper varies and as a result an emf is induced in the Loop
During the short interval of time where velocity of the jumper can be treated as constant ,the instantaneous value of the induced emf is

$E= -\frac {d\phi}{dt}=-vBa \cos \theta$ --(1)

Now current in the circuit at any point of time is given by
$I= \frac {dq}{dt}$

Where dq is the charge stored in the capacitor during the time dt

Now
q=CV
dq=CdV
or $dq= CBa \cos \theta dv$

Therefore Current at any time
$I= \frac {CBa \cos \theta dv}{dt}=CBa \cos \theta \frac {dv}{dt}$ ---(2)

Now Applying Newton's law on Jumper
$m \frac {dv}{dt}= mg \sin \theta - IaB \cos \theta $ ---(2)

Here $IaBcos \theta$ is the magnetic force acting on the jumper due to current
And Current will flow in such a way to oppose the motion

Substituting the value of I from equation (2) in equation (3)
$m \frac {dv}{dt}= mg \sin \theta - (CBa \cos \theta \frac {dv}{dt}) aB \cos \theta $
Or
$\frac {dv}{dt}=\frac {mg \sin \theta}{m + Ca^2B^2 \cos ^2 \theta}$

This is the acceleration of the Jumper

Now Time during which the jumper reach the bottom is given by
$b= \frac {1}{2} \frac {dv}{dt} t^2$
Or
$t_f=\sqrt { \frac {2b(m + Ca^2B^2 \cos ^2 \theta)}{mg \sin \theta}}$

Now velocity as a function of time is given by
$v= u + \frac {dv}{dt} t$

Or
$v=\frac {mg \sin \theta}{m + Ca^2B^2 \cos ^2 \theta} t$

Now induced EMF is given by
$E= -\frac {d\phi}{dt}=-vBa \cos \theta$
Or
$E=-\frac {(mg \sin \theta)(Ba \cos \theta)}{m + Ca^2B^2 \cos ^2 \theta} t$
Now velocity at the bottom is given
$v = \frac {dv}{dt} t_f = \sqrt {\frac {2bmg \sin \theta}{m + Ca^2B^2 \cos ^2 \theta}}$

Let us assume a small strip of width dx at a distance x from the wire
Magnetic field at distance x is given by
$B=\frac{\mu_0I}{2\pi x}$

So, flux through this small strip
$d\varphi=\frac{\mu_0I}{2\pi x}bdx$
Total flux will be given by

$\varphi=\int_{s}^{s+b}{\frac{\mu_0Ib}{2\pi x}dx=\frac{\mu_0Ib}{2\pi}ln{\frac{s+b}{s}}}$

$E=-\frac{d\varphi}{dt}=-\frac{d}{dt}[\frac{\mu_0Ib}{2\pi}ln{\frac{s+b}{s}}]$
Or
$=-\frac{\mu_0Ib}{2\pi}[\frac{1}{s+b}\frac{ds}{dt}-\frac{1}{s}\frac{ds}{dt}]$
$=-\frac{\mu_0Ib^2v}{2\pi b(s+b)}$

As no flux changes in this case,So induced EMF would be zero

The only 1.0m side have induced voltages
Let the side at y=.50 m be 1
Then
Induced emf on side 1
$E=BvL=.80e^{-.5 \times.5}(250)(1)=155.8$V

Induced emf on side 2
$E=BvL=.80e^{-.5 \times .6}(250)(1)=148.2$V
Both the EMF induced will have same polarity. The instantaneous current
$i=\frac{155.8-148.2}{2.5}=3.04$A

Velocity after falling from height h
$v=\sqrt{2gh}$

Induced EMF
$E=BL\sqrt{2gh}$

Induced current will be given by
$I=\frac{BL\sqrt{2gh}}{R}$

The force due to current
$F=BIL=\frac{B^2L^2\sqrt{2gh}}{R}$

The loop will attain terminal velocity if this force equals mg i.e if
$\frac{B^2L^2\sqrt{2gh}}{R}=Mg$

Or
$h=\frac{M^2gR^2}{2B^4L^4}$

Now
$E=-\frac{d\varphi}{dt}$

$\varphi=B.A=(B_0sin{2}\pi ft)(\pi r^2)$

So
$E=-2\pi^2r^2fB_0cos{2}\pi ft$

Resistance of the small dotted ring

$R=\frac{\rho L}{A}=\frac{\rho2\pi r}{wdr}$

Current is given by

$I=\frac{E}{R}=\frac{\pi f B_0w(cos{2}\pi ft)rdr}{\rho}$

For small ring
$dP=I^2R$
Or
$dP=\frac{2\pi^3B_0^2f^2w{cos}^2{2}\pi ft}{\rho}r^3dr$
Total power
$P=\int_{0}^{b}{dP=\frac{\pi^3wb^4f^2B_0^2}{2\rho}}{cos}^2{2}\pi ft$

Outside the solenoid ,magnetic field is zero and it is along the axis inside the solenoid
And it is given by
$B=\mu_0nI$

So flux through the wire
$\varphi=B.A=(\mu_0nI)(\pi r^2)=\mu_0\pi nI_0r^2sin{\omega}t$
Induced EMF
$E=-\frac{d\varphi}{dt}=-\mu_0\pi\omega nI_0r^2cos{\omega}t$

Induced Electric field
$E=\frac{InducedEMF}{2\pi R}=\frac{\mu_0\omega n I_0r^2cos{\omega}t}{2R}$

Motional emf =vBdl and since the rod is rotated in anticlockwise direction,As per lenz rule
For the part OB, polarity will O(+) and B(-)
For the part AO, polarity will O(+) and A(-)

Now Assume a small element dx at a distance x from origin
Induced EMF in the region
$e=\omega xB_0dx$
Total EMF for the region
$E_{OB}=\int_{0}^{b}{\omega x B_0dx=\frac{1}{2}B_0\omega b^2}$

Similary,
$E_{AO}=\frac{1}{2}B_0\omega a^2$
Taking polarity under consideration
$V_O-V_B=\frac{1}{2}B_0\omega b^2$
$V_O-V_A=\frac{1}{2}B_0\omega a^2$
So
$V_A-V_B=\frac{1}{2}B_0\omega(b^2-a^2)$

Let us assume a small strip of width dx at a distance x from the wire
Magnetic field at distance x is given by
$B=\frac{\mu_0I}{2\pi x}$

So flux through this small strip
$d\varphi=\frac{\mu_0I}{2\pi x}adx$
Total flux will be given by

$\varphi=\int_{a}^{2a}{\frac{\mu_0Ia}{2\pi x}dx=\frac{\mu_0Ia}{2\pi}ln{2}}$

Now induced EMF
$E=-\frac{d\varphi}{dt}=-\frac{\mu_0aln{2}}{2\pi}\frac{dI}{dt}=\frac{\mu_0aln{2}}{2\pi}\alpha$

Now induced current

$I=\frac{E}{R}=\frac{\mu_0a\alpha l n{2}}{2\pi R}$

$\frac{dQ}{dt}=\frac{\mu_0a\alpha l n{2}}{2\pi R}$
Or
$Q=\int_{0}^{1/\alpha}{\frac{\mu_0a\alpha l n{2}}{2\pi R}dt=\frac{\mu_0aln{2}}{2\pi R}}$

The emf of induction in the loop ABCD
$E_1=B_0b^2$

Similary the emf of induction in the loop BEFC
$E_2=\frac {B_0b^2}{2}$

These EMF will acts with accordance to the Lenz rule and will oppose change in flux
The circuit can be arranged with galvanic cell as the EMF of the induction as below

One the basis of ohm's law
$I_3br=E_1-I_13br$ --(1) $I_3br=I_22br-E_2$ --(2)

Also applying kirchoff law at the junction

$I_1=I_2+I_3$ --(3)

Solving equation (1), (2) and (3)
We get
$I_1=\frac{6E_1+2E_2}{22br}$
$I_2=\frac{2E_1+8E_2}{22br}$

$I_3=\frac{2E_1-3E_2}{11br}$

Substituting the values of $E_1$ and $E_2$ from above
$I_1=\frac{7B_0b}{22r}$
$I_2=\frac{3B_0b}{11r}$
$I_3=\frac{B_0b}{22r}$

The resistance of the Larger loop
$=2\pi R \times 10^{-4}=2\pi \times 1 \times 10^{-4}=2\pi10^{-4}$
Current in the larger loop will be
$I=\frac{V}{R}=\frac{(4+2.5t)}{2\pi10^{-4}}$

The magnetic Field due to this current in the larger loop at the common center will be
$B=\frac{\mu_0I}{2R}=\frac{\mu_0(4+2.5t)}{2*1*2\pi10^{-4}}=\frac{\mu_0(4+2.5t)}{4\pi10^{-4}}$

Since the smaller loop is very small compared to bigger loop,the magnetic field at the center can be assumed to be the magnetic field in the smaller loop
So flux through smaller loop
$\varphi=BA=\frac{\mu_0(4+2.5t)}{4\pi10^{-4}}(\pi10^{-2)}=\frac{\mu_0(4+2.5t)}{4}10^2$

Induced EMF in the smaller loop
$E=\frac{d\varphi}{dt}=62.5\mu_0$

Now resistance of the smaller loop
$=2\pi R*10^{-4}=2\pi*.1*10^{-4}=2\pi10^{-5}$
So induced current $I=\frac{E}{R}=\frac{62.5\mu_0}{2\pi10^{-5}}=1.25A$

Magnetic Field due to coil on the axis is given by
$B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}$

So Magnetic field at a distance z from coil PQR will be
$B=\frac{\mu_0Ia^2}{2(a^2+z^2)^{3/2}}$

Since the size of the coil ABC is very small,it may be assumed that magnetic field through it is uniform and equal to that on the axis

So flux through ABC
$\varphi=B.\pi b^2=\frac{\mu_0\pi I a^2b^2}{2(a^2+z^2)^{3/2}}$

Induced EMF
$E=\frac{d\varphi}{dt}=\frac{\mu_0I\pi a^2b^2}{2}\frac{d}{dt}\frac{1}{(a^2+z^2)^{3/2}}$
$=\frac{3\mu_0\pi I a^2b^2}{4(a^2+z^2)^{5/2}}2z\frac{dz}{dt}=\frac{3\mu_0\pi I a^2b^2zv}{2(a^2+z^2)^{5/2}}$