Electromagnetic Induction Test

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Electromagnetic Induction Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

When a current of 5 A flows in the primary coil then the flux linked with the secondary coil is 200 weber. The value of coefficient of mutual induction will be

A
1000 H

B
40 H

C
195 H

D
205 H

Solution

Coefficient of mutual induction will be the ratio of the flux linked with the secondary coil and the current primary coil.

$$ M = \dfrac{200}{5} = 40 H $$
Question 2
1 / -0

A straight copper wire is moved in a uniform magnetic field such that it cuts the magnetic lines of force. Then

A
emf will not be induced

B
emf will be induced

C
sometimes emf will be induced and sometimes not

D
nothing can be predicted

Solution

As the copper wire cuts the magnetic lines of force, the free charges in the wire will experience a force and will get displaced. This displacement of charges in the wire creates a potential difference and hence emf is induced.
Question 3
1 / -0

The coefficient of mutual induction between two coils is $$1.25 H.$$ If the rate of fall of current in the primary is $$80$$ As$$^{-1}$$, then the induced emf in the secondary coil will be

A
$$100 V$$

B
$$64 V$$

C
$$12.5 V$$

D
$$0.016 V$$

Solution

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

$$ = -1.25 \times -80= 100 V $$
Question 4
1 / -0

The rate of change of magnetic flux density through a circular coil of area 10 m and number of turns 100 is 10$$^3$$ Wb/m$$^2$$/s. The value of induced emf will be

A
10$$^{-2}$$ V

B
10$$^{-3}$$ V

C
10 V

D
10$$^{6}$$ V

Solution

$$emf =$$ rate of change of flux
$$=$$ area $$\times $$ number of turns $$\times $$ rate of change of flux density
$$= 10 \times 100 \times 1000 \\ =10^6 V $$
Question 5
1 / -0

A coil of area 80 cm$$^2$$ and number of turns 50 is rotating about an axis perpendicular to a magnetic field of 0.05 Tesla at 2000 rotations per minute. The maximum value of emf induced in it will be

A
$$200 \pi $$ volt

B
$$\displaystyle \frac{10 \pi}{3}$$ volt

C
$$\displaystyle \frac{4 \pi}{3}$$ volt

D
$$\displaystyle \frac{2}{3}$$ volt

Solution

$$Max. emf =n A B \omega $$

$$= 50 \times 0.008 \times 0.05 \times 2000 \dfrac{2 \pi}{60} = \dfrac{4 \pi }{3} V $$
Question 6
1 / -0

If the turns ratio of a transformer is 2 and the impedance of primary coil is 250 W then the impedance of secondary coil will be

A
1000 $$\Omega$$

B
500 $$\Omega$$

C
250 $$\Omega$$

D
125 $$\Omega$$

Solution

As shown in the image $$ L \propto N^2 $$
and transformer has coils which cause impedance and impedance of a inductor is $$2 \pi fL$$
so impedance of each coil $$ \propto N^2 $$
so impedance of secondary coil $$={ Z }_{ p }{ (\dfrac { { N }_{ S } }{ { N }_{ P } } })^{ 2 }\\=250*2^2 = 1000 \Omega$$
Question 7
1 / -0

The coefficient of self induction of a coil is given by

A
$$\displaystyle L = \left ( -\frac{dI}{dt} \right )$$

B
$$\displaystyle L = - \frac{edI}{dt}$$

C
$$\displaystyle L = \frac{dI}{edt}$$

D
$$\displaystyle L = \frac{dI}{dt}e^2$$

Solution

Inductance is the property of a conductor by which a change in current flowing through it induces (creates) a voltage (electromotive force) in both the conductor itself (self-inductance) and in any nearby conductors (mutual inductance). By Lenz's law the induced voltage opposed the the change in current. Hence inductance is defined as
$$L = - \dfrac{dI}{dt} $$
Question 8
1 / -0

Two circular conducting loops of radii $$R_1$$ and $$R_2$$ are laying concentrically in the same plane. If $$R_1$$ > $$R_2$$ then the mutual inductance (M) between them will be proportional to :

A
$$\displaystyle \frac{R_1}{R_2}$$

B
$$\displaystyle \frac{R_2}{R_1}$$

C
$$\displaystyle \frac{R_1}{R_2^2}$$

D
$$\displaystyle \frac{R_2^2}{R_1}$$

Solution

Let a current $$I_1$$ flows through.the outer circular coil of radius $$R_1$$.
The magnetic field at the centre of the coil is
$$B_1=\dfrac{μ_0I_1}{2R_1}$$
As the inner coil of radius $$R_2$$ placed co-axially has small radius $$(R_2<R_1)$$, therefore $$B_1$$ may be taken constant over its cross-sectional area.
Hence, flux associated with the inner coil is
$$\phi_2=B_1πR_2^2=\dfrac{μ_0I_1}{2R_1}πR_2^2$$

As, Mutual Inducatance, $$M=\dfrac{ϕ_2}{I_1}=\dfrac{μ_0πR_2^2}{2R_1}$$

∴$$M\propto \dfrac{R_2^2}{R_1}$$
Question 9
1 / -0

The length of side of a square coil is 50 cm and number of turns in it is 100. If it is placed at right angles to a magnetic field which is changing at the rate of 4 Tesla/s, then induced emf in the coil will be :

A
0.1 V

B
1.0 V

C
10 V

D
100 V

Solution

emf =|e|= $$\dfrac{d\phi}{dt}$$=NA$$\dfrac{dB}{dt}$$
rate of change of flux = $$N \times area \times $$ rate of change of magnetic field = $$100 \times 0.5^2 \times 4 = 100 V $$
Question 10
1 / -0

If a spark is produced on removing the load from an AC circuit then the element connected in the circuit is

A
high resistance

B
high capacitance

C
high inductance

D
high impedance

Solution

On removal of load from the circuit, the circuit suddenly becomes an open circuit.
Thus $$ \dfrac{di}{dt} \rightarrow \infty $$
For sparking, high voltage must appear across the open ends. This will happen only in case of an inductor as the voltage drop across the inductor is $$ L\dfrac{di}{dt} $$
Therefore, the circuit has high inductance.

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Re-Attempt Weekly Quiz Competition

Electromagnetic Induction Test - 31

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Electromagnetic Induction Test - 31

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SHARING IS CARING

Question 1

1000 H

40 H

195 H

205 H

Solution

Question 2

emf will not be induced

emf will be induced

sometimes emf will be induced and sometimes not

nothing can be predicted

Solution

Question 3

$$100 V$$

$$64 V$$

$$12.5 V$$

$$0.016 V$$

Solution

Question 4

10$$^{-2}$$ V

10$$^{-3}$$ V

10 V

10$$^{6}$$ V

Solution

Question 5

$$200 \pi $$ volt

$$\displaystyle \frac{10 \pi}{3}$$ volt

$$\displaystyle \frac{4 \pi}{3}$$ volt

$$\displaystyle \frac{2}{3}$$ volt

Solution

Question 6

1000 $$\Omega$$

500 $$\Omega$$

250 $$\Omega$$

125 $$\Omega$$

Solution

Question 7

$$\displaystyle L = \left ( -\frac{dI}{dt} \right )$$

$$\displaystyle L = - \frac{edI}{dt}$$

$$\displaystyle L = \frac{dI}{edt}$$

$$\displaystyle L = \frac{dI}{dt}e^2$$

Solution

Question 8

$$\displaystyle \frac{R_1}{R_2}$$

$$\displaystyle \frac{R_2}{R_1}$$

$$\displaystyle \frac{R_1}{R_2^2}$$

$$\displaystyle \frac{R_2^2}{R_1}$$

Solution

Question 9

0.1 V

1.0 V

10 V

100 V

Solution

Question 10

high resistance

high capacitance

high inductance

high impedance

Solution

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