Electromagnetic Induction Test

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Electromagnetic Induction Test - 39

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Question 1
1 / -0

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:

A
$$16mH$$

B
$$12mH$$

C
$$8mH$$

D
$$4mH$$

Solution

$$\displaystyle \therefore 45^o$$ phase angle means,
$$X_L=R$$
$$\therefore (2\pi fL)=R$$
$$\therefore L=\displaystyle\frac{R}{2\pi f}$$
$$\displaystyle =\frac{100}{(2\pi)(10^3)}$$
$$=0.0159H$$
$$=16mH$$
Question 2
1 / -0

Study involving both electricity and magnetism is called ______.

A
electromagnetism

B
magnetoelectricism

C
electricmagnetism

D
magneticelectromerism

Solution

Study of phenomenon involving both electricity and magnetism is called electromagnetism.
Question 3
1 / -0

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.

A
The induced emf between $$O$$ and $$Q$$ is $$Brv$$

B
An induced current $$I=\cfrac{2Bvr}{R}$$ flows in the clockwise direction

C
An induced current $$I=\cfrac{2Bvr}{R}$$ flows in the anticlockwise direction

D
No current flows

E
answer required

Solution

Since the magnetic field is uniform and normal to the plane of loop always, there is no change in flux through the ring due to its translation and rotation.
Thus there is no emf induced, and thus no current.
Question 4
1 / -0

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:

A
The entire rod is at the same electric potential

B
There is an electric field in the rod

C
The electric potential is highest at the center of the rod and decrease toward its ends

D
The electric potential is lowest at the center of the rod and increases toward its ends

Solution

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.
Question 5
1 / -0

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

A
zero

B
$$2{B}_{0}l{v}_{0}$$

C
$${B}_{0}l{v}_{0}$$

D
$$\sqrt {2} {B}_{0}l{v}_{0}$$

Solution

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, $$ e = l B.V $$
$$ = l B0 ( + ).V0( - ) $$
$$ = l B0 V0( + ).( - ) $$
$$ = l B0 V0(1-1) $$
$$ e   = 0   $$
Question 6
1 / -0

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 :

A
8V

B
4V

C
10V

D
6V

Solution

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.
Question 7
1 / -0

Constant magnetic field in the coil induces _________ current in it.

A
high

B
low

C
no

D
alternating

Solution

A current is induced in the coil if varying magnetic field lines pass through the coil. However, if the magnetic field is constant, no current is induced in it.
Question 8
1 / -0

According to Faraday's law, the total charge induced in a conductor that is moved in a magnetic field depends upon:

A
Initial magnetic flux

B
Final magnetic flux

C
Rate of change of magnetic flux

D
Average magnetic flux

Solution

According to Faraday's Law of induction,
total charge induced depends on the rate of change in magnetic flux .

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

And, $$Q\propto i\propto E$$

$$\implies Q\propto \dfrac{d\phi}{dt}$$

Answer-(C)
Question 9
1 / -0

A semi circular loop of radius R is placed in a uniform magnetic field as shown. It is pulled with a constantvelocity. The induced emf in the loop is

A
Bv ($$\pi$$ R) cos $$ \theta $$

B
Bv ($$\pi$$ R) sin $$ \theta $$

C
Bv (2R) cos $$ \theta $$

D
Bv (2R) sin $$ \theta $$

Solution

Join $$AB$$ and then also make a segment $$A'B'$$.
There is no change in flux in the whole semicircular loop, thus no emf is generated in it. But a certain emf is induced in equivalent segment $$A'B'$$ which is same as in between $$A$$ and $$B$$.
Length of $$A'B'$$, $$L = 2 R$$
$$\therefore$$ EMF induced between $$A'$$ and $$B'$$, $$\mathcal{E} = Bv_{\perp}L$$
$$\mathcal{E} = B (v sin\theta) (2R)$$
Question 10
1 / -0

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 ?

A
Moving the south pole in

B
Pulling the magnet out

C
Pushing the magnet in faster

D
Unwinding some of the turns of wire

Solution

When a magnet is moved closer to the current carrying coil it will generate electricity as the coil moves through the magnetic field. As the magnet is moved, there will be an induced electro-motive force (EMF) which can cause a current in the coil. Once the magnet stops moving, the current will go to zero.
Hence, when a galvanometer is connected to the circuit, there will be deflection due to the flow of electricity. As the magnet is moved toward the coil of wire, the needle of the galvanometer moves one direction. As the magnet is moved away from the coil of wire, the needle of the galvanometer moves the opposite direction. If the magnet is moved faster, the magnitude of the deflection increases.
Hence, pushing the magnet in faster will increase the size of the reading.

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Electromagnetic Induction Test - 39

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Electromagnetic Induction Test - 39

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Question 1

$$16mH$$

$$12mH$$

$$8mH$$

$$4mH$$

Solution

Question 2

electromagnetism

magnetoelectricism

electricmagnetism

magneticelectromerism

Solution

Question 3

The induced emf between $$O$$ and $$Q$$ is $$Brv$$

An induced current $$I=\cfrac{2Bvr}{R}$$ flows in the clockwise direction

An induced current $$I=\cfrac{2Bvr}{R}$$ flows in the anticlockwise direction

No current flows

answer required

Solution

Question 4

The entire rod is at the same electric potential

There is an electric field in the rod

The electric potential is highest at the center of the rod and decrease toward its ends

The electric potential is lowest at the center of the rod and increases toward its ends

Solution

Question 5

zero

$$2{B}_{0}l{v}_{0}$$

$${B}_{0}l{v}_{0}$$

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

Solution

Question 6

8V

4V

10V

6V

Solution

Question 7

high

low

no

alternating

Solution

Question 8

Initial magnetic flux

Final magnetic flux

Rate of change of magnetic flux

Average magnetic flux

Solution

Question 9

Bv ($$\pi$$ R) cos $$ \theta $$

Bv ($$\pi$$ R) sin $$ \theta $$

Bv (2R) cos $$ \theta $$

Bv (2R) sin $$ \theta $$

Solution

Question 10

Moving the south pole in

Pulling the magnet out

Pushing the magnet in faster

Unwinding some of the turns of wire

Solution

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