Electromagnetic Induction Test

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Electromagnetic Induction Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

The SI unit of inductance, the henry, can be written as :

A
Weber ampere$$^{-1}$$

B
Volt - s ampere$$^{-1}$$

C
Joule ampere$$^{-1}$$

D
ohm $$s^{-1}$$

Solution

$$V=L\dfrac{di}{dt}$$
So the unit $$L=\dfrac{Vdt}{di}$$
so SI unit of L is $$Volt-s\,ampere^{-1}$$
Question 2
1 / -0

Two coils of self inductances $$6mH$$ and $$8mH$$ are connected in series and are adjusted for highest co-efficient of coupling. Equivalent self inductance $$L$$ for the assembly is approximately

A
$$50mH$$

B
$$36mH$$

C
$$28mH$$

D
$$18mH$$

Solution

For highest co-efficient of coupling, the equivalent resistance of the assembly is given by:
$$L_{eq}=L_1+L_2+2\sqrt{L_1L_2}=6+8+2\sqrt(6\times8)=28mH$$
Question 3
1 / -0

An aircraft with a wingspan of $$40\ m$$ flies with a speed of $$1080\ km/hr$$ in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of the earth's magnetic field $$1.75\times 10^{-5}T$$. Then the emf developed between the tips of the wings is

A
$$0.5\ V$$

B
$$0.34\ V$$

C
$$0.21\ V$$

D
$$2.1\ V$$

Solution

Given : $$B = 1.75\times 10^{-5} T$$ $$l = 40 m$$
Speed of aircraft $$v = 1080$$ km/h $$ = 1080\times \dfrac{5}{18} =300$$ $$m/s$$
Emf induced $$\mathcal{E} = Bvl$$
$$\therefore$$ $$\mathcal{E} = (1.75\times 10^{-5}) (300)(40) = 0.21$$ volts
Question 4
1 / -0

While keeping area of cross-section of a solenoid same, the number of turns and length of solenoid one both doubled. The self inductance of the coil will be

A
Halved

B
Doubled

C
$$\dfrac{1}{4}$$ times the original value

D
Unaffected

Solution

The self inductance of a coil is $$L=N\dfrac{\phi}{I}$$ where $$N=$$ number of turns on coil, $$A=$$ cross section area and $$I=$$ current passes through coil.
Now, $$\phi=BA=(\dfrac{\mu_0NI}{l})A$$ where $$B=$$ magnetic field inside the solenoid and $$l=$$ length of the solenoid.
Thus, $$L=\mu_0\dfrac{N^2A}{l}$$
So, $$L'=\mu_0\dfrac{(2N)^2A}{(2l)}=2\mu_0\dfrac{N^2A}{l}=2L$$
Question 5
1 / -0

The self inductance of a coil having 500 turns is 50 mH. The magnetic flux through the cross-sectional area of the coil while current through it is 8 mA is found to be

A
$$4\times 10^{-4} Wb$$

B
$$0.04 Wb$$

C
$$4 \mu Wb$$

D
$$40 mWb$$

Solution

The magnetic flux through an inductor is the self inductance of coil times the current through it.
Thus $$\phi=LI$$
$$=50\times 10^{-3}H\times 8\times 10^{-3}A$$
$$=4\times 10^{-4}Wb$$
Question 6
1 / -0

A solenoid $$30 cm$$ long is made by winding $$2000$$ loops of wire on an iron rod whose cross-section is $$1.5{ cm }^{ 2 }$$. If the relative permeability of the iron is $$6000$$, what is the self-inductance of the solenoid?

A
$$15 H$$

B
$$2.5 H$$

C
$$3.5 H$$

D
$$0.5 H$$

Solution

Given : $$\mu_r = 6000$$ $$N = 2000$$ turns $$l = 0.3m$$
Cross-section area of solenoid $$A = 1.5\times 10^{-4}m^2$$
Self inductance $$L = \dfrac{\mu_r\mu_o N^2 A}{l}$$ where $$\mu = \mu_r \mu_o$$
$$\therefore$$ $$L = \dfrac{6000\times 4\pi \times 10^{-7} \times (2000)^2 \times 1.5\times 10^{-4}}{0.3} = 15 H$$
Question 7
1 / -0

In figure when key is pressed the ammeter A reads i ampere. The charge passing in the galvanometer circuit of total resistance R is Q. The mutual inductance of the two coils is :

A
$$\cfrac{Q}{R}$$

B
$$QR$$

C
$$\cfrac{QR}{i}$$

D
$$\cfrac{i}{QR}$$

Solution
Question 8
1 / -0

A metal disc of radius R rotates with an angular velocity, $$\omega = 1 rad/s$$ about an axis perpendicular to its plane passing through its centre in a magnetic field of induction B acting perpendicular to the plane of the disc. The induced e.m.f. between the rim and axis of the disc is:

A
$$BR^2$$

B
$$ 2B^2R^2$$

C
$$BR^3$$

D
$$BR^2/2$$

Solution

As shown in figure 1.AB is the axis of rotation, $$\vec { B } $$ is directed into the plane of disc consider a ring between the disc of thickness $$dx$$ at a distance $$x$$ from centre 0. The velocity of a ring $$V=wx$$ where w = angular velocity of a disc.
Now emf induced due to this ring is
$$de=VBdx\\ =Bwxd\\ \int _{ 0 }^{ e }{ de } =\int _{ 0 }^{ R }{ Bxdx } \\ e=B\left[ \dfrac { { x }^{ 2 } }{ 2 } \right] ^{ R }_{ 0 }\\ e=\dfrac { B{ R }^{ 2 } }{ 2 } $$as w= 1
Question 9
1 / -0

If emf induced in a coil is $$2V$$ by changing the current in it from $$8A$$ to $$6A$$ in $$2\times {10}^{-3}s$$, then the coefficient of self induction is

A
$$2\times { 10 }^{ -3 }H\quad $$

B
$$ { 10 }^{ -3 }H\quad $$

C
$$0.5\times { 10 }^{ -3 }H\quad $$

D
$$4\times { 10 }^{ -3 }H\quad $$

Solution

Given : $$\mathcal{E} = 2$$ V $$\Delta t = 2\times 10^{-3}$$ s
Change in current $$\Delta I = 6-8 = -2$$ A
Using $$\mathcal{E} = -L\dfrac{\Delta I}{\Delta t}$$

$$\therefore$$ $$2 = -L\times \dfrac{-2}{2\times 10^{-3}}$$ $$\implies L = 2\times 10^{-3}H$$
Question 10
1 / -0

Direction of induced EMF can be found from

A
Lenz law

B
Laplace law

C
Fleming law

D
None of the above

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Electromagnetic Induction Test - 45

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Electromagnetic Induction Test - 45

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Question 1

Weber ampere$$^{-1}$$

Volt - s ampere$$^{-1}$$

Joule ampere$$^{-1}$$

ohm $$s^{-1}$$

Solution

Question 2

$$50mH$$

$$36mH$$

$$28mH$$

$$18mH$$

Solution

Question 3

$$0.5\ V$$

$$0.34\ V$$

$$0.21\ V$$

$$2.1\ V$$

Solution

Question 4

Halved

Doubled

$$\dfrac{1}{4}$$ times the original value

Unaffected

Solution

Question 5

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

$$0.04 Wb$$

$$4 \mu Wb$$

$$40 mWb$$

Solution

Question 6

$$15 H$$

$$2.5 H$$

$$3.5 H$$

$$0.5 H$$

Solution

Question 7

$$\cfrac{Q}{R}$$

$$QR$$

$$\cfrac{QR}{i}$$

$$\cfrac{i}{QR}$$

Solution

Question 8

$$BR^2$$

$$ 2B^2R^2$$

$$BR^3$$

$$BR^2/2$$

Solution

Question 9

$$2\times { 10 }^{ -3 }H\quad $$

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

$$0.5\times { 10 }^{ -3 }H\quad $$

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

Solution

Question 10

Lenz law

Laplace law

Fleming law

None of the above

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