Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect but Reason is correct

MCQ Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 5

When the current in a coil changes from 2 amp. to 4 amp. in 0.05 sec., an e.m.f. of 8 volt induced in the coil. The coefficient of self inductance of the coil is

0%
0.1 henry
0%
0.2 henry
0%
0.4 henry
0%
0.8 henry

Explanation

Induced e.m.f.

$\epsilon=\dfrac {d\phi}{dt}$

$=\dfrac {dBA}{dt}=A_0\dfrac {dB}{dt}$

$=A_0\left (\dfrac {4B_0-B_0}{t}\right )=3A_0B_0/t$

A copper disc of radius

$0.1$

$m$ rotated about its centre with

$10$ revolutions per second in a uniform magnetic field of

$0.1$ tesla with it's plane perpendicular to the field. The e.m.f. induced across the radius of disc is:

0%
$\dfrac {\pi}{10}volt$
0%
$\dfrac {2\pi}{10}volt$
0%
$\pi \times 10^{-2}volt$
0%
$2\pi \times 10^{-2}volt$

Explanation

Consider a small radial segment of

$dx$ at a distance

$x$ from the centre of the disc. Velocity of this segment is

$\omega x$ . Emf induced across this segment

$de=vBL=\omega xB dx$
Total emf induced across radius of disc

$r$

$E=\dfrac{1}{2}B\omega r^2$

$\omega=2\pi n$ where

$n$ is number of revolutions per second

$B=0.1 T, r=0.1 m, \omega=10 rev/s=20\pi rad/s$

$E=\dfrac 12 \times 0.1 \times 0.01 \times 20\pi=\pi \times 10^{-2} V$

Magnetic flux

$\phi$ , in weber, in a closed circuit of resistance

$10\Omega$ varies with time

$t$ in seconds as

$\phi=6t^2-5t+1$ . The magnitude of induced current at

$t=0.25 s$ is :

0%
0.2 A
0%
0.6 A
0%
1.2 A
0%
0.8 A

Explanation

$e=\dfrac {-d\phi}{dt}=\dfrac {-d}{dt}(6t^2-5t+1)=-12t+5$

$e=-12(0.25)+5=2 volt$

$i=\dfrac {e}{R}=\dfrac {2}{10}=0.2 A$ .

A cylindrical bar magnet is kept along the axis of a circular coil. On rotating the magnet about its axis, the coil will have induced in it

0%
a current
0%
no current
0%
only an e.m.f.
0%
both an e.m.f. and a current

The coefficient of mutual inductance between two coils depends on

0%
medium between the coils
0%
separation between the two coils
0%
orientation of the two coils
0%
all of the above

A solenoid that is highly wound with wire of diameter

$0.10cm$ has a cross-sectional area

$0.90{cm}^{2}$ and is

$40cm$ long. Calculate the self inductance of the solenoid.

0%
$4.5 \times 10^{-5} H$
0%
$7.5 \times 10^{-5} H$
0%
$10 \times 10^{-5} H$
0%
$12.5 \times 10^{-5} H$

Explanation

Self-inductance in solenoid,

$L=\cfrac { { \mu }_{ 0 }{ N }^{ 2 }S }{ l }$

$and \quad N=\cfrac { l }{ d }$

$\Rightarrow \quad L=\cfrac { { \mu }_{ 0 }lS }{ { d }^{ 2 } } =\cfrac { \left( 4\pi \times { 10 }^{ -7 } \right) \left( 0.4 \right) \left( 0.9\times { 10 }^{ -4 } \right) }{ { \left( 0.1\times { 10 }^{ -2 } \right) }^{ 2 } } \quad =4.5\times { 10 }^{ -5 }H$

A horizontal telegraph wire

$0.5\ km$ long running east and west in a part of a circuit whose resistance is

$2.5\ \Omega$ . The wire falls to

$g=10.0\ m/s^2$ and

$B=2\times 10^{-5} weber/m^2$ , then the current induced in the circuit is

0%
$0.7\ amp$
0%
$0.04\ amp$
0%
$0.02\ amp$
0%
$0.01\ amp$

Explanation

$i=\dfrac {\varepsilon}{R}=\dfrac {1}{R}\dfrac {d\phi}{dt}$

Here

$df=B\times A=(2\times 10^{-5})\times 0.5\times 10^{+3}\times 5)$

$dt=$ time taken by the wire to fall at ground

$=(2h/g)^{1/2}=(10/10)^{1/2}=1 sec$

$\therefore i=\dfrac {1}{2.5}\left [\dfrac {(2\times 10^{-5})\times (0.5\times 10^3\times 5)}{1}\right ]=0.02 amp$ .

Which of the following units denotes the dimension

$\dfrac {ML^2}{Q^2}$ where Q denotes the electric charge?

0%
$Wb/m^2$
0%
$henry (H)$
0%
$H/m^2$
0%
$weber (Wb)$

Explanation

Mutual inductance

$=\dfrac {\phi}{I}=\dfrac {BA}{I}$

$[Henry]=\dfrac {[MT^{-1}Q^{-1}L^2]}{[QT^{-1}]}=ML^2Q^{-2}$

If coefficient of self induction of a coil is 1 H, an e.m.f. of 1 V is induced, if

0%
current flowing is 1 A
0%
current variation rate is 1 $As^{-1}$
0%
current of 1A flows for one sec
0%
none of these

Explanation

Induced EMF

$E =-L\dfrac{di}{dt}$ ...(i)

Given

$E=1\ V$ ,

$L=1 \ H$

$\Rightarrow \dfrac{di}{dt} =1 A/s$

The two rails of a railway track, insulated from each other and the ground, are connected to millivoltmeter. What is the reading of the millivoltmeter when a train passes at a speed of 180 km/hr along the track, given that the vertical component of earth's magnetic field is

$0.2\times 10^{-4}wb/m^2$ and rails are separated by 1 metre

0%
$10^{-1} volt$
0%
$10 \ mV$
0%
$1 \ volt$
0%
$1 \ mV$

Explanation

$\varepsilon =Blv$

$=(0.2\times 10^{-4})(1)(180\times 5/18)$

$=10^{-3}V$

$=1mV$

A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth's magnetic field is

$0.2\times 10^{-4}T$ , then the e.m.f developed between the two ends of the conductor is

0%
5 mV
0%
$50\mu V$
0%
$5\mu V$
0%
50 mV

Explanation

$I=2m, w=5 rad/s,$

$B=0.2\times 10^{-4}T$

$\varepsilon=\dfrac {B\omega l}{2}$

$=\dfrac {0.2\times 10^{-4}\times 5\times 1}{2}$

$=50\mu V$

The flux linked with a coil at any instant 't' is given by

$\phi=10t^2-50t+250$ . The induced emf at

$t=3s$ is

0%
-190 V
0%
-10 V
0%
10 V
0%
190 V

Explanation

Farady's law states that time varying magnetic flux can induce an e.m.f.
E = Electric field induced

$E = - \cfrac{d \phi}{dt}$

$E(t) = -\cfrac{d (10t^2 -50t+250)}{dt} = -(20t-50) V$

$E(t=3 s) = -(20(3)-50)V = -10 V$

The current in a coil of

$L=40 mH$ is to be increased uniformly from

$1A$ to

$11A$ in

$4$ milliseconds. The induced e.m.f. will be

0%
$100 V$
0%
$0.4V$
0%
$440 V$
0%
$40V$

Explanation

$e=\dfrac {LdI}{dt}$

$=\dfrac {40\times 10^{-3}(11-1)}{4\times 10^{-3}}$

$=100V$

The dimensions of self inductance are

0%
$\left[ ML{ T }^{ -2 }{ A }^{ -2 } \right]$
0%
$\left[ M{ L }^{ 2 }{ T }^{ -1 }{ A }^{ -2 } \right]$
0%
$\left[ M{ L }^{ 2 }{ T }^{ -2 }{ A }^{ -2 } \right]$
0%
$\left[ M{ L }^{ 2 }{ T }^{ -1 }{ A }^{ - 1} \right]$

Explanation

$U=\cfrac { 1 }{ 2 } L{ i }^{ 2 }\quad$

$\therefore \quad \left[ L \right] =\left[ \cfrac { U }{ { i }^{ 2 } } \right] =\left[ \cfrac { M{ L }^{ 2 }{ T }^{ -2 } }{ { A }^{ 2 } } \right]$

$=\left[ M{ L }^{ 2 }{ T }^{ -2 }{ A }^{ -2 } \right]$

A conducting rod is rotating about one end in a plane perpendicular to a uniform magnetic field with constant angular velocity. The correct graph between the induced emf(

$e$ ) across the rod and time (

$t$ ) is

Explanation

$EMF\quad induced\quad is\quad \frac { 1 }{ 2 } B\omega { l }^{ 2 },\quad which\quad is\quad constant\quad with\quad time.$

A solenoid that is highly wound with wire of diameter

$0.10cm$ has a cross-sectional area

$0.90{cm}^{2}$ and is

$40cm$ long. If the current through the solenoid decreases uniformly from

$10A$ to

$0A$ in

$0.10s$ , what is the emf induced between the ends of the solenoid?

0%
$4.5 \times 10^{-4} V$
0%
$5.5 \times 10^{-3} V$
0%
$5.0 \times 10^{-8} V$
0%
$4.5 \times 10^{-3} V$

Explanation

Induced emf,

$\left| e \right| =\left| L.\cfrac { \Delta i }{ \Delta t } \right|$

$=\cfrac { \left( 4.5\times { 10 }^{ -5 } \right) \left( 10 \right) }{ 0.1 }$

$=4.5\times { 10 }^{ -3 }V$

In figure, if the current

$i$ decreases at a rate

$\alpha$ , then

${V}_{A}-{V}_{B}$ is

0%
zero
0%
$-\alpha L$
0%
$\alpha L$
0%
No relation exists

Explanation

For an inductor

$\Delta V = L \cfrac{di}{dt}$
Given

$\cfrac{di}{dt}= - \alpha$

$V_A-V_B = - \alpha L$

The current

$i$ in an induction coil varies with time

$t$ according to the graph shown in the figure. Which of the following graphs shows the induced emf (

$\varepsilon$ ) in the coil with time?

Explanation

When the current is constant, the induced emf is zero as

$emf=-\dfrac { Ldi }{ dt }$ and when the current decreases at constant rate the emf is a positive constant and when it increase at constant rate it is a negative constant.

In figure, there is a conducting ring having resistance

$R$ placed in the plane of paper in a uniform magnetic field

${B}_{0}$ . If the ring is rotating in the plane of paper about an axis passing through point

$O$ and perpendicular to the plane of paper with constant angular speed

$\omega$ in clockwise direction, then

0%
point $O$ will be at higher potential than $A$
0%
the potential of point $B$ and $C$ will different
0%
the current in the ring will be zero
0%
the current in the ring will be $2{B}_{0}\omega {r}^{2}/R$
0%
answer required

Explanation

The ring can be seen as two rod OCA and OBA in parallel.

The motional emf will be induced between O and A =

$\dfrac{1}{2} B (2r)^2 \omega = 2Br^2 \omega$ with A positibe wrt O

In both the paths from O - C - A ; and from O - B - A , potential increases as one moves from point O to point A.
Point B and C are at same potential due to symmetry.
As there is no return path, no current flows even though a potential is developed between points O and A

Which of the following statement best describe the Faraday's Law?

0%
the induced voltage in a coil is proportional to the number of turns in the coil and to the rate at which the magnetic field is changing
0%
the induced voltage in a coil is equal to the rate at which the magnetic field is changing
0%
the induced voltage in a coil is proportional to the number of turns in the coil and to the strength of the magnetic field
0%
the induced voltage in a coil is proportional to the number of turns in the coil and to the size of the magnetic field
0%
the induced voltage in a coil proportional to the number of turns in the coil and to the source of the magnetic field

Explanation

Faraday's Law states that when the magnetic flux linking a circuit changes, an electromotive force is induced in the circuit proportional to the rate of change of the flux linkage.

Flux through a coil is given as

$\phi=(NA)B$

Thus the induced voltage is

$\dfrac{d\phi}{dt}=NA\dfrac{dB}{dt}$

$\propto N\dfrac{dB}{dt}$

An equilateral triangular conducting frame is rotated with angular velocity

$\omega$ in uniform magnetic field

$B$ as shown. Side of triangle is

$l$ . Choose the correct options

0%
${V}_{a}-{V}_{c}=0$
0%
${V}_{a}-{V}_{c}=\cfrac { B\omega { l }^{ 2 } }{ 2 }$
0%
${V}_{a}-{V}_{b}=\cfrac { B\omega { l }^{ 2 } }{ 2 }$
0%
${V}_{c}-{V}_{b}=-\cfrac { B\omega { l }^{ 2 } }{ 2 }$

Explanation

Velocity of each element between points a and c has components both along the element and perpendicular to it. However the direction of perpendicular components of velocity reverses for elements in the other half of the segment ac. Thus, the direction of potential difference created reverses for other half of the segment ac with equal magnitude. Hence,

$V_a=V_c$ .

Consider small elements of length

$dl$ in the segment ab. Velocity of each element perpendicular to the element

$=I\omega$

Thus,

$V_a-V_b=\int^{l}_0 B(l\omega)dl=\dfrac{1}{2}B\omega l^2$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

$\cfrac{di}{dt}=2A/s$

${V}_{a}-{V}_{b}=2A/s$

${V}_{a}-{V}_{b}=L\cfrac{di}{dt}$

$=(2)(2)=4V$

A metal disc of radius

$a$ rotates with a constant angular velocity

$\omega$ about its axis. The potential difference between the center and the rim of the disc is (

$m=$ mass of electron,

$e=$ charge on electron)

0%
$\cfrac{m{\omega}^{2}{a}^{2}}{e}$
0%
$\cfrac { 1 }{ 2 } \cfrac{m{\omega}^{2}{a}^{2}}{e}$
0%
$\cfrac{e{\omega}^{2}{a}^{2}}{2m}$
0%
$\cfrac{e{\omega}^{2}{a}^{2}}{m}$

Explanation

For second law of Motion,

$F=ma$

$eE=mw^2r$

$E=\dfrac{mw^2r}{e}$

The potential difference

$V=\int E.dr$

$V=\dfrac{\int_{0}^{a}mw^2r}{e}=\dfrac{mw^2a^2}{2e}$

A metal rod of resistance

$20\Omega$ is fixed along a diameter of a conducting ring of radius

$0.1m$ and lies on x-y plane. There is a magnetic field

$\vec { B } =\left( 50T \right) \hat { K }$ . The ring rotates with an angular velocity

$\omega=20rad$

${s}^{-1}$ about its axis. An external resistance of

$10\Omega$ is connected across the center of the ring and rim. The current through external resistance is:

0%
$\cfrac{1}{4}$
0%
$\cfrac{1}{2}$
0%
$\cfrac{1}{3}$
0%
zero

Explanation

The potential difference is

$V=\dfrac{1Bwr^2}{2}=\dfrac{1*20*50(0.1)^2}{2}=5V$

The equivalent circuit is

$i=\dfrac{V}{R_1+R_2}=\dfrac{5}{10+5}=\dfrac{1}{3}A$

Two identical cycle wheels (geometrically) have different number of spokes connected from center to rim. One is having

$20$ spokes and the other having only

$10$ (the rim and the spokes are resistanceless). One resistance of value

$R$ is connected between center and rim. The current in

$R$ will be

0%
double in the first wheel than in the second wheel
0%
four times in the first wheel than in the second wheel
0%
will be double in the second wheel than that of the first wheel
0%
will be equal in both these wheels
0%
answer required

A uniform magnetic field exists in a region given by

$\vec { B } =3\hat { i } +4\hat { j } +5\hat { k }$ . A rod of length

$5m$ along y-axis moves with a constant speed of

$1m$

${s}^{-1}$ along x axis. Then the induced emf in the rod will be

0%
$0$
0%
$25V$
0%
$20V$
0%
$15V$
0%
none

Explanation

$Given, B = 3 + 4 + 5k$

$V = 1 m/s (As \ the \ rod \ moves \ in \ X-Direction)$

$Length \ of \ rod = 5 m$

$Since, e = -d\phi /dt \ and \ \phi = B.A$

$So, e = -d(B.A )/dt$

$e = d(B.A)/dt ( Negative \ sign \ is \ neglected \ as \ it \ only \ represents \ the \ direction)$

$e = B.d(A)/dt ( Magnetic \ Field \ is \ constant )$

$= B.d(l\times b)/dt ( Where \ l = Length \ and \ b = Breadth )$

$= l B. d(b)/dt ---- (1)$

As $db/dt$ is nothing but the velocity with which the rod is moving.

$So, db/dt = V$

Putting the above value of $db/dt$ in equation (1)
We have, $e = l B.V$

But, in the above equation it is assumed that the direction of B, L and V are mutually perpendicular.

So, Only that component of B that is in mutual perpendicular direction of L and V will be taken into account.

$So, e = l B.V = 5 \times 5 \times 1$

$e = 25 V$

When an

$AC$ signal of frequency

$1kHz$ is applied across a coil of resistance

$100\Omega$ , then the applied voltage leads the current by

$45^o$ . The inductance of the coil is:

0%
$16mH$
0%
$12mH$
0%
$8mH$
0%
$4mH$

Explanation

$\displaystyle \therefore 45^o$ phase angle means,

$X_L=R$

$\therefore (2\pi fL)=R$

$\therefore L=\displaystyle\frac{R}{2\pi f}$

$\displaystyle =\frac{100}{(2\pi)(10^3)}$

$=0.0159H$

$=16mH$

A rectangular coil

$ABCD$ is rotated anticlockwise with a uniform angular velocity about the axis shown in figure. The axis of rotation of the coil as well as the magnetic field

$B$ is horizontal. The induced emf in the coil would be minimum when the plane of the coil

0%
is horizontal
0%
makes and angle ${45}^{o}$ with direction of magnetic field
0%
is at right angle to the magnetic field
0%
makes and angle of ${30}^{o}$ with the magnetic field
0%
answer required

Explanation

Let the angle between magnetic field and the line perpendicular to the surface of rectangular coil be

$\theta$ .

Hence the flux through the coil is

$\phi=\vec{B}.\vec{A}=BAcos\theta$

Hence the induced emf=Rate of change of flux=

$-\dfrac{d\phi}{dt}=BAsin\theta \dfrac{d\theta}{dt}=BA\omega sin\theta$

Hence it is minimum when

$\theta=0^{\circ},180^{\circ}$

Which means when the plane of coil is at right angle to the magnetic field.

Study involving both electricity and magnetism is called ______.

0%
electromagnetism
0%
magnetoelectricism
0%
electricmagnetism
0%
magneticelectromerism

A conducting ring of radius

$r$ and resistance

$R$ rolls on a horizontal surface with constant velocity

$v$ . The magnetic field

$B$ is uniform and is normal to the plane of the loop. Choose the correct option.

0%
The induced emf between $O$ and $Q$ is $Brv$
0%
An induced current $I=\cfrac{2Bvr}{R}$ flows in the clockwise direction
0%
An induced current $I=\cfrac{2Bvr}{R}$ flows in the anticlockwise direction
0%
No current flows
0%
answer required

A metal rod moves at a constant velocity in a direction perpendicular to its length. A constant uniform magnetic field exists in space in a direction perpendicular to the rod as well as its velocity. Select the correct statement(s) from the following:

0%
The entire rod is at the same electric potential
0%
There is an electric field in the rod
0%
The electric potential is highest at the center of the rod and decrease toward its ends
0%
The electric potential is lowest at the center of the rod and increases toward its ends

Explanation

Hint:

Whenever any charge moves in a Magnetic Field, a force acts on it whose direction is given by Fleming Left Hand Rule.

The correct Option is B.

The explanation for the correct answer:

$\bullet$ Fleming Left Hand Rule states that if we stretch the thumb, forefinger, and middle finger of our left hand such that they are perpendicular to each other. Then if the middle finger represents the direction of motion of positive charge and the forefinger represents the direction of the magnetic field, then the thumb gives the direction of force acting on the charge.

$\bullet$ In the given question, the rod moves in a direction perpendicular to its length in a magnetic field which is perpendicular to the length of the rod as well as with the direction of the motion of the rod. Then force acts on the negative and positive charges in the rod.

$\bullet$ By using the Fleming Left-Hand rule, the direction of the force on negative and positive charges is such that all the negative charges get collected at the one end of the rod and all the positive charges to the other end of the rod.

$\bullet$ The end with positive charges is at higher potential whereas the end with negative charges is at lower potential. A potential difference is created at the ends of the rod, due to which an electric field is produced in the rod. Thus, Option B is correct and Option A is incorrect.

$\bullet$ The highest potential in the rod is at one end of the rod that contains positive charges, and the lowest potential to the other end contains negative charges. Thus Option C and D are incorrect.

Thus, only Option B is correct.

A conductor

$AB$ of length

$l$ moves in x-y plane with velocity

$\vec { v } ={ v }_{ 0 }\left( \hat { i } -\hat { j } \right)$ . A magnetic field

$\vec { B } ={ B }_{ 0 }\left( \hat { i } +\hat { j } \right)$ exists in the region. The induced emf is

0%
zero
0%
$2{B}_{0}l{v}_{0}$
0%
${B}_{0}l{v}_{0}$
0%
$\sqrt {2} {B}_{0}l{v}_{0}$

Explanation

Given, $B = B0 ( + ) ; V = V0 ( - )$

Since,

$e = -d\phi /dt and \phi = B.A$

So,

$e = -d(B.A )/dt$

$e = d(B.A)/dt$ (Negative sign is neglected as it only represents the direction)

$e = B.d(A)/dt$ (Magnetic field is constant)

$= B.d(lb)/dt$ (Where l = Length and b = Breadth)

$= l B. d(b)/dt$ ---- (1)

As,

$db/dt$ is nothing but the velocity with which the rod is moving.

So,

$\dfrac{db}{dt} = V$

Putting the above value of

$db/dt$ in equation (1).

We have,

$e = l B.V$

$= l B0 ( + ).V0( - )$

$= l B0 V0( + ).( - )$

$= l B0 V0(1-1)$

$e = 0$

The conductor

$AD$ moves to the right in a uniform magnetic field directed into the plane of the paper.

0%
The free electron in $AD$ will move toward $A$
0%
$D$ d will acquire a positive potential with respect to $A$
0%
$A$ current will flow from $A$ to $D$ in $AD$ in closed loop
0%
The current in $AD$ flows from lower to higher potential
0%
answer required

Explanation

$E=( v \times B).l$

Therefore inside the rod, the electric field will have the direction from A to D.

Therefore negatively charged electron will move towards A.

This will cause the potential of A to decrease with respect to D.

Positive charge will flow from A to D along the electric field. So the direction of current in AD in closed loop is from A to D.

As D is at a higher potential, current in AD flows from lower(A) to higher potential(D)

A vertical conducting ring of radius

$R$ falls vertically in a horizontal magnetic field of magnitude

$B$ . The direction of

$B$ is perpendicular to the plane of the ring. When the speed of the ring is

$v$ ,

0%
no current flows in the ring
0%
$A$ and $D$ are at the same potential
0%
$C$ and $E$ are at the same potential
0%
the potential difference between $A$ and $D$ is $2BRv$ , with $D$ at a higher potential
0%
answer required

The magnetic flux through a circuit of resistance 'R' changes by an amount

$\Delta \Phi$ at a time

$\Delta t$ . Then the total quantity of electric charge Q that passes any point in the circuit during the time

$\Delta t$ is represented by

0%
$Q = \displaystyle \frac{\Delta \Phi}{R}$
0%
$Q= \displaystyle \frac{\Delta \Phi}{\Delta t}$
0%
$Q= \displaystyle R. \frac{\Delta \Phi}{\Delta t}$
0%
$Q = \displaystyle {1}{R} . \frac{\Delta \Phi}{\Delta t}$

Explanation

$\displaystyle I = \frac{e}{R} = \frac{1}{R} \left | \frac{\Delta \phi}{\Delta t} \right | = \frac{\Delta Q}{\Delta t}$

$\displaystyle Q = \frac{\Delta \phi}{R}$

Comment on the statement given below:

In self-induction

When the current in a coil is increasing, induced emf opposes it
When the current in a coil is decreasing, induced emf supports it

0%
A is true, B is false
0%
A and B are false
0%
A and B are true
0%
A is false, B is true

Which of the following statement is correct regarding induced electric field (symbols have their usual meanings)?

0%
Work done in moving a test charge in an induced electric field can be zero
0%
Induced electric field is non-conservative is nature
0%
Induced electric lines of force form closed loops
0%
Induced e.m.f. in the loop in $\displaystyle \varepsilon =\oint \overrightarrow{E}.\overline{d\iota }=-\frac{d\phi }{dt}$

Explanation

The induced emf is perpendicular to the motion of particle hence, work done is zero

Hence,

$A$ is the answer

Induced electric force is non conservative in nature, from the definiton

Hence,

$B$ is the answer

Induced E.M.F are formed in closed circles

Hence,

$C$ is the answer

From the definition, we have

$\int { E.dl= -d\phi /dt }$

hence

$D$ is also the answer

A magnetic field of flux density 1.0 Wb

$m^{-2}$ acts normal to a 80 tum coil of 0.01

$m^2$ area. The e.m.f. induced in it, if this coil is removed from the field in 0.1 second is :

0%
8V
0%
4V
0%
10V
0%
6V

Explanation

Answer is A.

The magnetic flux

$\phi =B\times A$ depends on the magnetic field, the area of the loop, and their relative orientation.
In this case, the magnetic field of flux density

$1.0Wb{m}^{-2}$ acts normal to a 80 turns coil of

$0.01{m}^{2}$ area.
Therefore, the Magnetic flux in 80 turns of coil = 0.01 * 80 = 0.8 T.
The time derivative of the flux is given as, d

$\phi$ /dt.
Therefore, the e.m.f induced in it, if this coil is removed from the field in 0.1 second is d

$\phi$ /dt = 0.8/0.1 = 8 V.
The e.m.f induced is 8 V.

A wire 88 cm long bent into a circular loop is placed perpendicular to the magnetic field of flux density 2.5 Wb

$m^{-2}$ . Within 0.5 s, the loop is changed into a square and flux density is increased to 3.0 Wb

$m^{-2}$ . The value of e.m.f. induced is :

0%
0.018V
0%
0.016V
0%
0.020V
0%
0.012V

Explanation

Answer is A.

The magnetic flux

$\phi =B\times A$ depends on the magnetic field, the area
of the loop, and their relative orientation.
In this case, the wire 88 cm long bent into a circular loop is placed perpendicular to the magnetic field and Within 0.5 s, the loop is changed into a square.
Magnetic Flux when in a circular loop:
Circumference of the loop =

$2\times 3.14\times r\quad =\quad 88\quad cm$ .
So, radius of the loop r = 14 cm.
Magnetic field

$\phi =B\times A\quad =2.5\times 3.14\times 0.14\times 0.14=0.154\quad Wb$ .
Magnetic Flux when in a square:
Circumference of the square =

$4r\quad =\quad 88\quad cm$
So, length of the side of the square is

$\frac { 88 }{ 4 } =\quad 22cm$ .
Magnetic feild

$\phi '=B\times A\quad =3\times 0.22\times 0.22=0.1452\quad Wb$ .
The total induced emf is

$\frac { 0.154-0.1452 }{ 0.5 } =0.018\quad V$ .
Therefore, the value of e.m.f induced is 0.018 V.

Which of the following electrical devices works on the principle of electro-magnetic induction ?

0%
Electric fan
0%
Electric bulb
0%
Electric cooker
0%
L.E.D.

A magnet is moved into the coil of wire as shown, there is a small reading on the sensitive meter.
Which change would increase the size of the reading ?

0%
Moving the south pole in
0%
Pulling the magnet out
0%
Pushing the magnet in faster
0%
Unwinding some of the turns of wire

The simplest type of AC voltage or current is the one which

0%
Varies exponentially
0%
Varies sinusoidally
0%
Varies linearly
0%
Does not vary uniformly

The current flowing through the primary coil of mutual inductance 8 H is reduced to zero in

$10^{-3}$ s. As a result of which a

$24\times 10^3$ V is induced in the secondary. The initial current through the primary is _____A.

0%
2
0%
1
0%
4
0%
3

Explanation

Given,

$M= 8 H$

$dI= I_2-I_1= -I_1A$

$dt= 10^{-3}s$

$\varepsilon= 24\times10^3 V$

The induced emf

$\varepsilon$ in the secondary coil is,

$\varepsilon= -M\dfrac{dI}{dt}$

$24\times 10^3=\dfrac{8\times I_1}{10^{-3}}$

$I_1= \dfrac{24}{8}A$

$I_1=3A$

What will be the magnitude of e.m.f. induced in a 200 turns coil with cross section area 0.16

$m^2$ if the magnetic field through the coil changes from 0.10 Wb

$m^{-2}$ to 0.30

$m^{-2}$ in 0.05 second.

0%
128V
0%
130V
0%
118V
0%
132V

Explanation

Answer is A.

The magnetic flux

$\phi =B\times A$ depends on the magnetic field, the area of the loop, and their relative orientation.
In this case, The magnetic field through the coil changes from

$0.10Wb{m}^{-2}$ to

$0.30Wb{m}^{-2}$ , at a uniform rate over a period of 0.05 s.
The change in magnetic feild is

$0.10Wb{m}^{-2}$ -

$0.30Wb{m}^{-2}$ =

$0.20Wb{m}^{-2}$ .
The magnetic flux in one turn of the coil =

$\phi =B\times A\quad =\quad 0.20\times 0.16\quad =\quad 0.032T$
Therefore, the Magnetic flux in 200 turns of coil = 0.032 * 200 = 6.4 T.
Therefore, the e.m.f induced in it, if this coil is removed from the field in 0.05 second is d

$\phi$ /dt = 6.4/0.05 = 128 V.
The magnitude of the e.m.f induced is 128 V.

The variation of anode current in a triode valve corresponding to a change in grid potential at three different values of the plate potential is shown in the given figure. The mutual conductance of triode is

0%
$5\times {10}^{-3} mho$
0%
$2.5\times {10}^{-3} mho$
0%
$7.5\times {10}^{-3} mho$
0%
$9.5\times {10}^{-3} mho$

Explanation

Mutual conductance

$=\cfrac{\Delta {I}_{p}}{\Delta {V}_{g}}=\cfrac{5\times {10}^{-3}}{2}$

$=2.5\times {10}^{-3}mho$

Consider a current carrying coil placed in a magnetic field. What are the requirements for induced current to flow as per electromagnetic induction.

0%
Coil of wire carrying current
0%
Change in magnetic field associated with the coil
0%
Both A and B
0%
None

When the coil and the magnet are both stationary

0%
there is no deflection in the galvanometer.
0%
galvanometer deflects.
0%
galvanometer bursts.
0%
none

A jet plane is travelling towards west at a speed of 1600 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earths magnetic field at the location has a magnitude of 5

$\times 10^{-4}$ T and the dip angle is

$30^o$ .

0%
4.1 V
0%
2.2 V
0%
3.2 V
0%
3.8 V

Explanation

Vertical component of B,

$B \sin \delta = 5 \times 10^{-4} \times \sin (30)=2.5 \times 10^{-4} T$

EMF,

$E=Blv$

$E=2.5 \times 10^{-4} \times 25 \times 1800 \times \dfrac{5}{18}=3.125 V$

If a current carrying coil is close to a magnet and both are moving with the same speed in same direction, what is the effect on induced current ?

0%
Induced current increases
0%
Induced current decreases
0%
Induced current flows to oppose motion
0%
Induced current remains equal to zero

What is the self inductance of a coil in which an induced emf of

$2V$ is set up, when the current is changing at the rate of

$4As^{-1}$

0%
$0.5\ mH$
0%
$0.05\ H$
0%
$2\ H$
0%
$0.5\ H$

Explanation

Induced emf in a inductor

$L$ is:

$\varepsilon = L \dfrac{di}{dt}= L \times 4$

$\Rightarrow 2 = L \times 4$

$\Rightarrow L = 0.5\: H$

CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 5 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 5 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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