Electromagnetic Induction Test

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Electromagnetic Induction Test - 51

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Question 1
1 / -0

Who gave the principle of Electro-magnetic induction?

A
Volta

B
Oerstead

C
Ampere

D
Faraday

Solution

The principle of Electro-magnetic induction was given by Michael Faraday in 1831.
Question 2
1 / -0

The equivalent quantity of mass in electricity is :

A
current

B
self inductance

C
potential

D
charge

Solution

B. self-inductance

The equivalent quantity of mass in electricity is self- inductance. Self-inductance of an inductor is referred to as the "inertia of electricity" .

Kinetic energy in Mechanics,
K= $$\dfrac{1}{2}mv^2$$
where m= mass of an object.
v= velocity of the object.

The magnetic energy stored in an inductor,
K'= $$\dfrac{1}{2}LI^2$$
where L= self-inductance of the inductor and,
I= current through the inductor.

So, from the above equation, we can conclude that self-inductance is equivalent to mass in electricity.
Question 3
1 / -0

Mutual inductance of two coils can be increased by

A
decreasing the number of turns in the coils

B
increasing the number of turns in the coils

C
winding the coils on wooden cores

D
none of these.

Solution

As M = $$\dfrac{\mu_0N_1N_2A}{l} $$
i.e M can be increased by increasing the numbers of turns in the coils.
Question 4
1 / -0

The self inductance of two solenoids A & B having equal length are same. If the number of turns in two solenoids A & B are $$100$$ and $$200$$ respectively. The ratio of radii of their cross-section will be __________.

A
$$2:1$$

B
$$1:2$$

C
$$1:4$$

D
$$4:1$$

Solution

Self inductance of a solenoids is given by $$\dfrac{\mu_0 N^2A}{l}$$.
Given. $$\dfrac{\mu_0 N_1^2A_1}{l_1}$$ = $$\dfrac{\mu_0 N_2^2A_2}{l_2}$$
$$\Rightarrow$$ $$\dfrac{100\times 100\mu_0 A_1}{l_1}$$ = $$\dfrac{ 200\times200\mu_0 A_2}{l_2}$$
$$\Rightarrow$$ $$A_1 = 4A_2(\because l_1 =l_2)$$
$$\Rightarrow$$ $$\dfrac{r_1}{r_2}= 2 $$
Therefore, correct answer is $$2$$.
Question 5
1 / -0

A 60 volt - 10 watt bulb is operated at 100 volt - 60 Hz a.c. The inductance required is?

A
2.56 H

B
0.32 H

C
0.64 H

D
1.28 H
Question 6
1 / -0

A magnetic flux of 5 microweber is linked with a coil when a current of 1 mA flows through it. Calculate self-inductance of the coil.

A
5 mA

B
10 mH

C
15 mH

D
20 mH

Solution

The correct answer is A
Magnetic flux $$B=5mw=5\times 10^{-6}mw$$

$$current I= 1mA=10^{-3}A$$

$$self-inductanceL=\frac{Q}{I}$$

$$L=\frac{5\times 10^6}{10^{-3}}$$

$$L=5\times 10^{-3}$$ Hen.

$$L=5mh$$
Question 7
1 / -0

A source of 220 V is applied in an A C circuit . The value of resistance is 220 $$\Omega$$. Frequency & inductance are 50Hz & 0.7 H then wattless current is

A
0.5 amp

B
0.7 amp

C
1.0 amp

D
None

Solution

A source= $$220V$$
The value of resistance= $$220 \Omega$$
Frequency= $$50 Hz$$
Inductance= $$0.7H$$
Find the wattless current= ?
Wattless component of current is $$i=i_v\sin \theta$$
$$=\cfrac {Ev}{z}\sin \theta$$
where, $$z=$$ impedance of $$L-R$$ circuit
$$=\sqrt {R^2+L^2W^2}$$ so,
$$i=\cfrac {220}{\sqrt {R^2+L^2+W^2}}\sin \theta$$ from impedance triangle,
$$\sin \theta= \cfrac {LW}{\sqrt {R^2+L^2W^2}}$$
$$\Rightarrow i=\cfrac {220}{\sqrt {R^2+L^2W^2}}\cfrac {LW}{\sqrt {R^2+L^2W^2}}$$
$$=\cfrac {220}{R^2+L^2W^2}LW$$
$$=\cfrac {220 \times 0.7 \times 2 \Pi \times 50}{(220)^2+(0.7\times 2\Pi \times 50)^2}$$
$$=\cfrac {220 \times 220}{(220)^2+(220)^2}$$
$$=0.5 A$$ .
Question 8
1 / -0

A small square loop of wire of side $$l$$ is placed inside a large square loop of wire of side $$L(L>>l)$$.
The loops are co-planar and their centres coincide. The mutual inductance of the system is proportional to:

A
$$l/L$$

B
$${l}^{2}/L$$

C
$$L/l$$

D
$${L}^{2}/l$$

Solution

The correct answer is option (B).
Hint: Calculate the magnetic flux and use it to find out the mutual inductance.

Step 1: Calculate magnetic flux of smaller loop.

Let the length of the larger square and the smaller square be $$L$$ and $$l$$ respectively. Accordingly, the current in these two square loops are $$I$$ and $$i$$.

According to Faraday’s first law of electromagnetic radiation, whenever magnetic flux passes through surface changes with respect to time, an induced EMF is generated.
Magnetic field at a distance $$\dfrac{L}{2}$$ from the centre of the current carrying wire of length $$L$$ is given by:

$$B_1=\dfrac{\mu _0}{4\pi}\dfrac{i}{\dfrac{L}{2}}2sin 45$$

$$B_1=\dfrac{\mu _0}{4\pi}\dfrac{i\times 2\times 2sin \times \dfrac{1}{\sqrt{2}}}{L}$$

Magnetic field in the centre of square due to four wires of length $$L$$ will be:

$$B_1=\dfrac{\mu _0}{4\pi}\dfrac{4\times i\times 2\times 2sin \times \dfrac{1}{\sqrt{2}}}{L}$$

$$B_1=\dfrac{\mu_0}{\pi}\dfrac{2\sqrt2i}{L}$$

Since, the smaller loop is at the centre of the larger loop, so $$L>>l$$

In a small square, the magnetic field is uniform and is equal to $$B_1$$

Flux in the smaller loop will be:

$$\phi =B_1l^2$$

$$\phi =\dfrac{\mu_0}{\pi}\dfrac{2\sqrt2il^2}{L}$$

Step 2: Calculate the mutual inductance.

Mutual inductance is given by –

$$M=\dfrac{\phi}{i}$$

$$M =\dfrac{\mu_0}{\pi}\dfrac{2\sqrt2l^2}{L}$$

$$M\propto \dfrac{l^2}{L}$$

Hence, the mutual inductance of the system is proportional to $$\dfrac{l^2}{L}$$.
The correct answer is option (B).
Question 9
1 / -0

A flat coil of 500 turns, each of area $$50cm^2$$, rotates in a uniform magnetic field of $$0.14 Wb/m^2$$ about an axis normal to the field at an angular speed of $$150 rad/s$$. The coil has a resistance of $$5\Omega$$. The induced e.m.f. is applied to an external resistance of $$10\Omega$$, The peak current through the resistance is:

A
1.5 A

B
2.5 A

C
3.5 A

D
4.5 A

Solution
Question 10
1 / -0

A circular coil of radius $$8$$cm and $$20$$ turns rotates about its vertical diameter with an angular speed of 50 rad $$s^{-1}$$ in a uniform horizontal magnetic field of $$8\times 10^{-2}T.$$ The maximum emf induced in the coil is nearly

A
6.3 V

B
0.6 V

C
1.6 V

D
0.15 V

Solution

The emf(e) induced in the coil rotating in magnetic field $$ \overrightarrow { B } $$ with an angular velocity $$\omega \dfrac{rad}{s}$$ is given as a function of time.
$$\varepsilon =NBA\omega\sin(\omega t)$$
where ,$$N$$=number of turn of coil
$$A$$=area of cross section
$$\varepsilon_{max}$$ when $$\sin\omega t=1$$ ,$$\omega t=90°$$
$$\varepsilon _{max}=NBA\omega=20\times8\times 10^{-2} \times 50\times \pi(8\times 10^{-2})^{2}$$
$$\varepsilon_{max}=1.6V$$

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Electromagnetic Induction Test - 51

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Electromagnetic Induction Test - 51

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Question 1

Volta

Oerstead

Ampere

Faraday

Solution

Question 2

current

self inductance

potential

charge

Solution

Question 3

decreasing the number of turns in the coils

increasing the number of turns in the coils

winding the coils on wooden cores

none of these.

Solution

Question 4

$$2:1$$

$$1:2$$

$$1:4$$

$$4:1$$

Solution

Question 5

2.56 H

0.32 H

0.64 H

1.28 H

Question 6

5 mA

10 mH

15 mH

20 mH

Solution

Question 7

0.5 amp

0.7 amp

1.0 amp

None

Solution

Question 8

$$l/L$$

$${l}^{2}/L$$

$$L/l$$

$${L}^{2}/l$$

Solution

Question 9

1.5 A

2.5 A

3.5 A

4.5 A

Solution

Question 10

6.3 V

0.6 V

1.6 V

0.15 V

Solution

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