Electromagnetic Induction Test

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Electromagnetic Induction Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

When a current of 2 A is passed through a coil of 100 turns, flux associated with it is $$5\, \times\, 10^{-5}$$ Wb.
Find the self inductance of the coil.

A
$$4\, \times\, 10^{-3}\, H$$

B
$$4\, \times\, 10^{-2}\, H$$

C
$$2.5\, \times\, 10^{-3}\, H$$

D
$$10^{-3}\, H$$

Solution

Total flux passing through a coil=$$N\phi=Li$$
where $$L$$ is the self inductance of coil.

$$\implies L=\dfrac{N\phi}{i}=\dfrac{100\times 5\times 10^{-5}}{2}$$

$$=2.5\times 10^{-3}H$$
Question 2
1 / -0

Voltage in the secondary coil of a transformer does not depend upon :

A
frequency of the source

B
voltage in the primary coil

C
ratio of number of turns in the two coils

D
both (b) and (c)

Solution

The voltage in primary and secondary coil is related as:
$$\Rightarrow \dfrac { { V }_{ S } }{ { V }_{ P } } =\dfrac { no.of\ turns\ in\ secondary\ coil\ { (N }_{ s })\ }{ no.of\ turns\ in\ primary\ coil\ ({ N }_{ p }) } $$

Clearly, voltage in secondary coil depends on voltage in primary coil and number of turns in primary and secondary coil. It is independent of frequency of the source.
Hence, option A is correct.
Question 3
1 / -0

A straight conductor of length $$0.4 m$$ is moved with a speed of $$7 {m}/{s}$$ perpendicular to the magnetic field of intensity of $$0.9 {Wb}/{{m}^{2}}$$. The induced emf across the conductor will be :

A
$$7.25 V$$

B
$$5.54 V$$

C
$$1.25 V$$

D
$$2.52 V$$

Solution
Question 4
1 / -0

A fighter plane of length 20 m, wing span (distance from tip of one wing to the tip of the other wing) of 15 m and height 5 m is flying towards east over Delhi. Its speed is $$240 ms^{-1}$$. The earth's magnetic field over Delhi is $$5 \times 10^{-5}$$ T with the declination angle $$\sim 0^o$$ and dip of $$\theta$$ such that $$sin \theta = \frac{2}{3}$$. If the voltage developed is $$V_B$$ between the lower and upper side of the plane and $$V_{W}$$ between the tips of the wings then $$V_{B}$$ and $$V_{W}$$ are close to

A
$$V_B= 40 mV; V_W= 135 mV$$ with left side of pilot at higher voltage

B
$$V_B =45 mV; V_W= 120 mV $$ with right side of pilot at higher voltage

C
$$V_B= 40 mV; V_W= 135 mV$$ with right side of pilot at high voltage

D
$$V_B = 45 mV; V_W= 120 mV$$ with left side of pilot at higher voltage

Solution

$$V_B=VB_Hl = 240\times 5\times 10^{-5}cos(\theta)\times 5$$
$$=44.7mv$$
By right hand rule, the change moves to the left of pilot
Question 5
1 / -0

The magnitude of the induced emf in a coil of inductance $$30\ mH$$ in which the current changes from $$6\ A$$ to $$2\ A$$ in $$2\ s$$ is:

A
$$0.06\ V$$

B
$$0.6\ V$$

C
$$1.06\ V$$

D
$$6\ V$$

Solution

Induced emf is given by:
$$e=-L\dfrac{di}{dt} $$
$$=-30\times 10^{-3}\times \dfrac{4}{2}$$
$$=0.06\ V$$
Question 6
1 / -0

A metallic wire bent in form of a semi-circle of radius 0.1 m is moved in a direction parallel to its plane, but perpendicular to a magnetic field $$B = 20$$ mT with a velocity of 10 m/s. What is the induced e.m.f. in wire?

A
$$4 \times 10^{-3}$$ volts

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

C
$$4 \times 10^{-1}$$ volts

D
None of these

Solution

Consider an infinitesimally small element of length $$Rd\theta$$ in the wire.
Net emf developed across the element=$$Bvsin\theta (Rd\theta)$$
Thus the total emf in the wire across two ends=$$\int_0^{\pi}vsin\theta BRd\theta$$
$$=2BvR$$
$$=2\times 0.02\times 10\times 0.1\ volts$$
$$=0.04\ volts$$
Question 7
1 / -0

A glass rod of length $$\ell$$ moves with a velocity $$v$$ in a uniform magnetic field $$B$$, what will be the emf induced in the rod:

A
$$Blv$$

B
$$\dfrac{Bl}{v}$$

C
$$Bl$$

D
None of these

Solution

Glass is an Insulator so no EMF will be induced.
Question 8
1 / -0

The magnetic flux linked with a coil at any instant $$t$$ is given by $$\displaystyle \phi =5{ t }^{ 3 }-100t+300$$, the emf induced in the coil at $$t=2\ s$$ is:

A
$$40\ V$$

B
$$-40\ V$$

C
$$300\ V$$

D
$$140\ V$$

Solution

Magnetic flux $$\phi = 5t^3 - 100t + 300$$
Differentiating w.r.t time t we get $$\dfrac{d\phi}{dt} = 15t^2 -100$$

$$\therefore$$ Emf induced at $$t = 2$$ s $$E = - \dfrac{d\phi}{dt} \bigg|_{t =2}$$
$$\therefore$$ $$E = - [15 \times (2)^2 -100] = 40$$ V
Question 9
1 / -0

A rectangular coil of $$300$$ turns has an average area of $$25\ cm \times 10\ cm$$. The coil rotates with a speed of $$50\ cps$$ in uniform magnetic field of strength $$4 \, \times \, 10^{-2} \, T$$ about an axis perpendicular to the field. The peak value of the induced emf is (in volt):

A
$$300 \, \pi$$

B
$$3000 \, \pi$$

C
$$3 \, \pi$$

D
$$30 \, \pi$$

Solution

Peak voltage will be induced when the longer side moves perpendicular to the magnetic field in which case the axis will be parallel to the longer end.

$$\omega =2\pi f=2\pi (50)=100\pi $$

Area $$=A=250\ cm^2=250 \times 10^{-4}\ m^2$$

Peak induced voltage will be:
$$V=NAB \omega$$
$$\implies300 \times 250 \times 10^{-4} \times 4 \times 10^{-2} \times 100 \pi$$

$$=3\pi\ volt$$

Question 10

1 / -0

Match the following:

Quantity	Formula
1) Magnetic flux linked with a coil	a) $$\displaystyle -N\frac { d\phi }{ dt } $$
2) Induced emf	b) $$\displaystyle { \mu }_{ r }{ \mu }_{ 0 }{ n }_{ 1 }{ n }_{ 2 }{ \pi r }_{ 1 }^{ 2 }l$$
3) Force on a charged particle moving in a electric and magnetic field	c) $$\displaystyle BA\cos { \theta } $$
4) Mutual inductance of a solenoid	d) $$\displaystyle q\left( \overline { E } +\overline { v } \times \overline { B } \right) $$

1-c, 2-d, 3-b, 4-a

1-c, 2-a, 3-d, 4-b

1-b, 2-a, 3-c, 4-d

1-a, 2-b, 3-d, 4-c

Solution

1) Magnetic flux through any area is the scalar product of its area vector with the magnetic field vector. Thus for a coil, it is $$\vec{B}.\vec{A}=BAcos\theta$$

2) Emf induced in a coil due to changing flux through it is given by Faraday's Law,

$$Emf = -N\dfrac{d\phi}{dt}$$

3) Force on a charged particle due to electric field = $$q\vec{E}$$

Force on a moving charged particle due to magnetic field = $$q(\vec{v}\times \vec{B})$$

Thus, force on a moving charged particle in an electric and magnetic field = $$q(\vec{E}+\vec{v}\times \vec{B})$$

4) Mutual inductance of a solenoid is found out to be : $$\mu_r\mu_0n_1n_2\pi r_1^2l$$

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Electromagnetic Induction Test - 43

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Electromagnetic Induction Test - 43

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Question 1

$$4\, \times\, 10^{-3}\, H$$

$$4\, \times\, 10^{-2}\, H$$

$$2.5\, \times\, 10^{-3}\, H$$

$$10^{-3}\, H$$

Solution

Question 2

frequency of the source

voltage in the primary coil

ratio of number of turns in the two coils

both (b) and (c)

Solution

Question 3

$$7.25 V$$

$$5.54 V$$

$$1.25 V$$

$$2.52 V$$

Solution

Question 4

$$V_B= 40 mV; V_W= 135 mV$$ with left side of pilot at higher voltage

$$V_B =45 mV; V_W= 120 mV $$ with right side of pilot at higher voltage

$$V_B= 40 mV; V_W= 135 mV$$ with right side of pilot at high voltage

$$V_B = 45 mV; V_W= 120 mV$$ with left side of pilot at higher voltage

Solution

Question 5

$$0.06\ V$$

$$0.6\ V$$

$$1.06\ V$$

$$6\ V$$

Solution

Question 6

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

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

$$4 \times 10^{-1}$$ volts

None of these

Solution

Question 7

$$Blv$$

$$\dfrac{Bl}{v}$$

$$Bl$$

None of these

Solution

Question 8

$$40\ V$$

$$-40\ V$$

$$300\ V$$

$$140\ V$$

Solution

Question 9

$$300 \, \pi$$

$$3000 \, \pi$$

$$3 \, \pi$$

$$30 \, \pi$$

Solution

Question 10

1-c, 2-d, 3-b, 4-a

1-c, 2-a, 3-d, 4-b

1-b, 2-a, 3-c, 4-d

1-a, 2-b, 3-d, 4-c

Solution

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