Electromagnetic Induction Test

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Electromagnetic Induction Test - 44

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Question 1
1 / -0

What will be the self inductance of a coil of 100 turns if a current of 5 A produces a magnetic flux $$5\times 10^{-5}$$ Wb?

A
1 mH

B
10 mH

C
1 $$\mu$$H

D
10 $$\mu$$H

Solution

Magnetic field inside the coil=$$\phi=Li$$

$$\implies L=\dfrac{\phi}{i}=\dfrac{5\times 10^{-5}}{5}H=10\mu H$$.
Question 2
1 / -0

The current in a coil changes from $$1 \ mA$$ to $$5 \ mA$$ in $$4$$ milli second. If the coefficient of self-induction of the coil is $$10\ mH$$ the magnitude of the "self-induced" emf is:

A
$$10\ mV$$

B
$$5\ mV$$

C
$$2.5 \ mV$$

D
$$1\ mV$$

Solution

Self induced emf, $$e=-L \dfrac{dI}{dt} $$
$$= 10 \times {10}^{-3} \times \dfrac{4 \times {10}^{-3}}{4 \times {10}^{-3}}=10\ mV$$
Question 3
1 / -0

An airplane with a wing span of $$40m$$ changes its pitch so that it is flying $$100m/s$$ in an area of the Earth's magnetic field that is perpendicular to the plane and equal to $$5.010^{5}T$$.
What is the emf induced between the wing tips? Assume that the plane was previously flying parallel to the earths magnetic field.

A
0.12 v

B
0.20 v

C
1.2 v

D
2.0 v

Solution

Given : $$l = 40$$ m $$v = 100$$ m/s $$B = 5.0 \times 10^{-5}$$ T
$$\therefore$$ Emf produced between the wing tips $$\mathcal{E} =Bvl$$
$$\implies$$ $$\mathcal{E} = 5.0\times 10^{-5} \times 100 \times 40 = 0.20$$ v
Question 4
1 / -0

A $$2 cm$$ long bar slides along metal rails at a speed of $$1 {cm}/{s}$$ towards the right side. The magnetic field inside the bar and rails is $$2 T$$, pointing out of the page. Find out the induced emf in the bar and rails?

A
$$2 \times {10}^{-5} V$$

B
$$2 \times {10}^{-4} V$$

C
$$4 \times {10}^{-4} V$$

D
$$2 \times {10}^{-3} V$$

E
$$4 \times {10}^{-3} V$$

Solution

Given : $$L =2 cm = 0.02$$ m $$v = 1 cm/s = 0.01$$ $$m/s$$ $$B =2T$$
$$\therefore$$ Induced emf $$\mathcal{E} = BvL = 2 \times 0.01 \times 0.02 = 4 \times 10^{-4}$$ $$V$$
Question 5
1 / -0

Identify which of the following best describe the Mutual inductance?

A
the ability of a current carrying conductor to induce a voltage in another conductor through a mutual magnetic field.

B
the ability of current carrying conductor to produce a changing magnetic field.

C
the ability of a conductor to induce a magnetic field in another current carrying conductor.

D
the ability of a current carrying conductor to induce a current in another conductor through a mutual magnetic field.

E
the ability of a magnetic field to induce a voltage in a current carrying conductor.

Solution

As per the Faraday's experiments, a current in a conductor produces a magnetic field. This magnetic field gets linked with another conductor and result in an induced emf.
Question 6
1 / -0

When a voltage source that is inducing voltage into a large number of coils is disconnected, and a switch that is in series with the coils of wire is also opened, a spark is observed to jump across the switch terminals as the switch begins to open up. Identify the cause of this spark?

A
Free electrons from the voltage source

B
Free electrons from the coils of wire

C
Collapse of the magnetic field in the coils

D
Secondary electron flow from the source

E
Stored voltage in the coils of wire

Solution

As the voltage is disconnected, there will be a decreases in the magnetic field of coils. Magnetic energy has to be transferred so, a spark is generated. thus, magnetic energy is released in the form of fire.
Question 7
1 / -0

Fill in the blank.
Electromagnetic induction occurs in a wire when a change occurs in ______ in wire.

A
current

B
intensity of the electric field

C
voltage applied

D
magnetic field intensity applied

E
resistance added to the wire

Solution

Electromagnetic induction occurs because of the change in the flux through the coil. Flux through the coil can be changed by changing the current or by magnetic field or by changing the area of the coil.
Question 8
1 / -0

As shown in above figure, a uniform magnetic field B, direction of magnetic field is outward of the plane of the page. Calculate the potential difference between point a and b if metal rod is pulled upward with constant velocity.

A
0

B
$$\dfrac{1}{2}vBL$$, with point a at the higher potential

C
$$\dfrac{1}{2}vBL$$, with point b at the higher potential

D
vBL, with point a at the higher potential

E
vBL, with point b at the higher potential

Solution

When the rod is moved upward with velocity $$v$$ each electron $$q$$ inside the rod experiences a Lorentz magnetic force
$$ {f}_{m}=Bqv$$ (given that magnetic field and velocity are perpendicular to each other)
it acts from point b to point a (by Fieming's left hand rule) ,due to this magnetic force electrons accumulate at the end a so end a becomes negative and end b positive.Hence an electric field is set up between the points b and a from b to a .The electric field so produced exerts electric force on electrons from a to b and opposes the magnetic Lorentz force. But due to accumulation of electrons at point a electric field increases hence electric force, after some time a stage comes when both forces become equal .If E is electric field at this stage then electric force on a electron
$${f}_{e}=qE$$
in equilibrium both forces will be equal and opposite
$${f}_{e}= $$$$ {f}_{m}$$
$$qE$$ =$$Bqv$$
$$E=Bv$$
if V be the potential difference and L is the length of rod , then
$$E=V/L$$
or we can say from last two equations ,
$$V/L=vB$$
$$V=vBL$$
As point is positive so it will be at higher potential.
Question 9
1 / -0

Magnitude of e.m.f produced in a coil, when a magnet is inserted into it does not depend upon

A
Number of turns in the coil.

B
Resistance of the coil.

C
Magnetic moment of magnet.

D
Speed of the magnet.

Solution

$$e=N\dfrac{(\phi_2-\phi_1)}{t}=\dfrac{N\Delta(BA)}{t}$$
So it is independent of magnetic moment of magnet. change in flux is due to velocity.
Question 10
1 / -0

An aeroplane is moving towards north horizontally with a speed of $$200 m/s$$ at a place where the vertical component of earths magnetic field is $$0.5\times { 10 }^{ -4 }$$ tesla. Then the induced e.m.f. set up between the tips of the wings of the plane if they are $$10 m$$ apart is:

A
0.1 volt

B
0.01 volt

C
10 volt

D
1 volt

Solution

$$EMF=Blv$$

$$=.5\times { 10 }^{ -4 }\times 10\times 200$$

$$=0.1V$$

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Electromagnetic Induction Test - 44

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Electromagnetic Induction Test - 44

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Question 1

1 mH

10 mH

1 $$\mu$$H

10 $$\mu$$H

Solution

Question 2

$$10\ mV$$

$$5\ mV$$

$$2.5 \ mV$$

$$1\ mV$$

Solution

Question 3

0.12 v

0.20 v

1.2 v

2.0 v

Solution

Question 4

$$2 \times {10}^{-5} V$$

$$2 \times {10}^{-4} V$$

$$4 \times {10}^{-4} V$$

$$2 \times {10}^{-3} V$$

$$4 \times {10}^{-3} V$$

Solution

Question 5

the ability of a current carrying conductor to induce a voltage in another conductor through a mutual magnetic field.

the ability of current carrying conductor to produce a changing magnetic field.

the ability of a conductor to induce a magnetic field in another current carrying conductor.

the ability of a current carrying conductor to induce a current in another conductor through a mutual magnetic field.

the ability of a magnetic field to induce a voltage in a current carrying conductor.

Solution

Question 6

Free electrons from the voltage source

Free electrons from the coils of wire

Collapse of the magnetic field in the coils

Secondary electron flow from the source

Stored voltage in the coils of wire

Solution

Question 7

current

intensity of the electric field

voltage applied

magnetic field intensity applied

resistance added to the wire

Solution

Question 8

0

$$\dfrac{1}{2}vBL$$, with point a at the higher potential

$$\dfrac{1}{2}vBL$$, with point b at the higher potential

vBL, with point a at the higher potential

vBL, with point b at the higher potential

Solution

Question 9

Number of turns in the coil.

Resistance of the coil.

Magnetic moment of magnet.

Speed of the magnet.

Solution

Question 10

0.1 volt

0.01 volt

10 volt

1 volt

Solution

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