What is the acceleration of the jumper
Question 2
What is the time taken to reach bottom
Question 3
Find the instantaneous EMF as a function of time t
Question 4
What will be the velocity at the bottom
(a)$ \sqrt {\frac {2bmg \sin \theta}{m + Ca^2B^2 \cos ^2 \theta}}$
(b)$ \sqrt {\frac {2bmg \cos \theta}{m + Ca^2B^2 \cos ^2 \theta}}$
(c)$ \sqrt {\frac {2amg}{m + Ca^2B^2 \cos ^2 \theta}}$
(d)$ \sqrt {\frac {2bmg}{m + Ca^2B^2 \cos ^2 \theta}}$
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
$E=\frac{d\varphi}{dt}=\frac{\mu_0I\pi a^2b^2}{2}\frac{d}{dt}\frac{1}{(a^2+z^2)^{3/2}}$
$=\frac{3\mu_0\pi I a^2b^2}{4(a^2+z^2)^{5/2}}2z\frac{dz}{dt}=\frac{3\mu_0\pi I a^2b^2zv}{2(a^2+z^2)^{5/2}}$
The frame will oscillate Simple harmonic motion.
The equation of simple harmonic motion is given by
$ x=Asin{(\omega t+\delta)}$
Assuming that frame is at its equilibrium position at time t=0
We have
$x=Asin(\omega t) $ ---(1)
Where
$\omega=\sqrt{\frac{g}{L}}$
Now the flux linked through wire frame at any instant is given by
$\varphi=BS cos{\alpha}$
Where $\alpha$ is the angle made by the wire at time t
$E=\frac{d\varphi}{dt}=BSsin{\alpha}\frac{d\alpha}{dt}$
Now as a is small we have $sin \alpha = \alpha$
Also $\alpha=\frac {x}{L}
So
$E=BS\left(\frac{x}{L}\right)\frac{d}{dt}\left(\frac{x}{L}\right)=\frac{BSx}{L^2}\frac{dx}{dt}$
From equation (1)
$E=\frac{BS}{L^2}Asin{\omega}t\frac{d}{dt}Asin{\omega}t=\frac{BSA^2\omega}{2L^2}sin{2}\omega t$
Let us take a small ring of width dr at a distance r from the center.
The resistance of the small ring is
$R=\rho\frac{2\pi r}{hdr}R$
The EMF induced in this small ring due to changing magnetic field
$E=\frac{d\varphi}{dt}=B_0\pi x^2$
So current in this ring
$I=\frac{E}{R}=\frac{B_0hxdx}{2\rho}$
Total current through entire ring
$I=\int_{a}^{b}\frac{B_0hxdx}{2\rho}=\frac{B_0h}{4\rho}(b^2-a^2)$
As x increase so $\frac {dB}{dt}$ increase i.e., induced emf (e) is negative. When loop completely entered in the magnetic field, emf=0
when it exit out x increase but $\frac {dB}{dt}$ decrease i.e., e is positive.
So correct option is (c)
Correct answer is (a)
Between the inner wire and sheath only the inner wire contributes to B.
And we know Magnetic field due to inner wire is given by
$B=\frac{\mu_0I}{2\pi r}$
We consider a rectangular area between the surface of the inner wire and outer sheath and parallel to cable axis. We break this area into long parallel strips of width dr and length ds. Flux through sdr is Bsdr and total flux between $r=R_1$ and $r=R_2$ per unit length is given by
$\varphi=\frac{1}{s}\int_{R_1}^{R_2}{\frac{\mu_0Is}{2\pi r}=\frac{\mu_0I}{2\pi}ln{\frac{R_2}{R_1}}}$
Now self inductance is given by
$L=\frac{\varphi}{I}=\frac{\mu_0}{2\pi}ln{\frac{R_2}{R_1}}$
Self inductance is given by
$I_iL_i=N_i\varphi_i$
v
So self inductance for coil X
$L=\frac{300*1.2*10^{-4}}{1.5}=24mH$
We cannot find self inductance of coil B as data is insufficient
Now
Mutual inductance is given by
$I_xM=N_y\varphi_y$
So M will be
$M=\frac{600*.9*10^{-4}}{1.5}=36mH$
The mutual inductance of the solenoid and coil system is given by
$M=\mu_0n_1N_2A_2$
Where $n_1$ -> Number of turns per unit length in outer solenoid
$N_2$ -> Number of turns in the small inner solenoid
$A_2$ -> Area of the small solenoid
So here in this case
Substituting all the values
$M=8\pi^2 \times 10^{-5}$H
Now change in current
$\Delta I=-2-2=-4$A
$\Delta t=.05$
Now induced Emf is given by
$E=-M\frac{dI}{dt}$
Substituting the values
$E=6.3 \times 10^{-2}$V
Therefore current in the coil will be given by
$I=\frac {E}{R}$
Substituing the values from above
$I=6.4 \times 10^{-4}$ A
Now
Flux in the small can be calculated using Mutual inductance
$\varphi=MI$
Or
$\varphi=2 \times 8\pi^2 \times 10^{-5}=16\pi^2 \times 10^{-5}$ Weber
We suppose that straight wire is carrying current I downward.
Now we know that Magnetic Field due to long straight wire at a distance x is given by
$B=\frac{\mu_0I}{2\pi x}$
Let us consider a small strip of width dx at a distance x from the long wire on the rectangular loop.
The flux through small strip will be given by
$d\varphi=\frac{\mu_0I}{2\pi x}adx$
Total flux through rectangular loop will be given by
$\varphi=\int_{c}^{c+b}{\frac{\mu_0I}{2\pi x}adx=\frac{\mu_0Ia}{2\pi}ln{(}1+\frac{b}{c})}$
Now mutual inductance is given by
$M=\frac{\varphi}{I}=\frac{\mu_0a}{2\pi}ln{(}1+\frac{b}{c})$
Since circular loop is pulled to make triangle,
Perimeter of Circle=Perimeter of Triangle
$\pi D=3a\bigma=\frac{\pi D}{3}$
Now Area of circular Loop
$A_1=\pi\left(\frac{D}{2}\right)^2=\frac{\pi D^2}{4}$
Area of equilateral triangle
$A_2=\frac{\sqrt3}{4}a^2=\frac{\sqrt3}{4}\left(\frac{\pi D}{3}\right)^2=\frac{\pi^2D^2\sqrt3}{36}$
Now induced Emf
$E=B(A_1-A_2)$
$E=\frac{B\pi D^2}{4}\left(1-\frac{\pi\sqrt3}{4}\right)$
Since flux does not change with time
No induced Emf generated.
So a is true
The magnetic Field at the distance s from long straight wire is given by
$B=\frac{\mu_0I}{2\pi s}$
Let us consider a small rectangular strip of width ds at a distance s from center on the rectangular crosssection of the toroid
Then
$d\varphi=\frac{\mu_0I}{2\pi s}hds$
Total flux
$\varphi=\int_{a}^{b}{\frac{\mu_0I}{2\pi s}hds=\frac{\mu_0Ih}{2\pi}ln{\frac{b}{a}}}$
Now we have N turns ,So total flux
$\varphi=\frac{\mu_0IhN}{2\pi}ln{\frac{b}{a}}$
Now $I=I_0cos{\omega}t$
So
$\varphi=\frac{\mu_0I_0(cos{\omega}t)hN}{2\pi}ln{\frac{b}{a}}$
Now EMF induced is
$E=-\frac{d\varphi}{dt}=\frac{\mu_0Nh\omega}{2\pi}ln{\left(\frac{b}{a}\right)}I_0sin{\omega}t$ ---(1)
The magnetic Field inside the toriod is given by
$B=\frac{\mu_0NI}{2\pi s}$
The flux through a single turn
$\varphi=\int_{a}^{b}{Bhds=\frac{\mu_0NIh}{2\pi}ln{\frac{b}{a}}}$
Total flux is N times,So self inductance would be
$L=\frac{\mu_0N^2h}{2\pi}ln{\frac{b}{a}}$
From equation (1) ,induced EMF
$E=-\frac{d\varphi}{dt}=\frac{\mu_0Nh\omega}{2\pi}ln{\left(\frac{b}{a}\right)}I_0sin{\omega}t$
So induced current in the toriod
$I=\frac{E}{R}=\frac{\mu_0Nh\omega}{2\pi R}ln{\left(\frac{b}{a}\right)}I_0sin{\omega}t$
Now this current I changing,So Back EMF generated will be
$E_b=-L\frac{dI}{dt}=-\frac{\mu_0N^2h}{2\pi}ln{\frac{b}{a}}\frac{\mu_0Nh\omega}{2\pi R}I_0sin{\omega}tln{\frac{b}{a}}$
$E_b=\frac{\mu_0^2N^3h^2\omega I_0}{4\pi^2R}\left(ln{\frac{b}{a}}\right)^2sin{\omega}t$
(a), (b), (c)
Correct option is (d)
Using $k_1$,$k_2$ etc, as different constants.
$I_1(t)=k_1[1-e^{-t/\tau}]$,$B(t)=k_2I_1(t)$
Now $I_2(t)$
$I_2(t)=k_3 \frac {dB(t)}{dt}=k_4 e^{-t/\tau}$
$I_2(t)B(t)=k_5[1-e^{-t/\tau}][e^{-t/\tau}]$
This quantity is zero for t=0 and t=$\infty$ and positive for other value of t. It must, therefore, pass through a maximum.
(b)