Electromagnetic Induction Test

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Electromagnetic Induction Test - 52

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Question 1
1 / -0

Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be

A
maximum in situation $$(a)$$

B
maximum in situation $$(b)$$

C
maximum in situation $$(c)$$

D
the same in all situations

Solution

Reffer image.

Mutual induction $$\propto $$ flux linkage between coils

1) flux linkage in coil (a) is maximum because both coils have same axis.

2) Flux linkage between coils is $$0$$ because thus axis are perpendicular to each other.

3) Flux linkage between coils is $$0$$ because their axis are perpendicular to each other .

So, mutual inductance between coils is maximum in case (a).
Question 2
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A conducting rod of length l is rotating with constant angular velocity $$\omega$$ about an axis O' in a uniform magnetic field B as shown in figure. The emf induced between ends P and Q will be (the axis is parallel to B)

A
$$\dfrac{1}{4}$$ B$$\omega l^2$$

B
$$\dfrac{5}{10}$$ B$$\omega l^2$$

C
zero

D
Non of the above.
Question 3
1 / -0

A simple pendulum with bob of mass $$m$$ and conducting wire of length $$L$$ swings under gravity through an angle $$2\theta$$. The earth's magnetic field component in the direction perpendicular to swing is $$B$$.The maximum potential difference induced across the pendulum is:

A
$$2BL\sin { \left( \cfrac { \theta }{ 2 } \right) } \sqrt { gL } $$

B
$$BL\sin { \left( \cfrac { \theta }{ 2 } \right) } \sqrt { gL } $$

C
$$BL\sin { \left( \cfrac { \theta }{ 2 } \right) } { \left( gL \right) }^{ 3/2 }$$

D
$$BL\sin { \left( \cfrac { \theta }{ 2 } \right) } { \left( gL \right) }^{ 2 }$$

Solution

We have $$h=L(1-Cos\theta)$$

Maximum velocity at equilibrium is given by,

$$v^2=2gh=2gL(1-Cos\theta)$$

$$=2gL(2Sin^2(\dfrac{\theta}{2}))$$

$$v=2\sqrt{gL}sin(\dfrac{\theta}{2})$$

Thus Maximum potential difference,

$$V_{max}=BvL$$

$$=B\times 2\sqrt{gL}sin(\dfrac{\theta}{2}) L$$

$$=2BLsin (\dfrac{\theta}{2}) \sqrt{gL}$$
Question 4
1 / -0

A metallic square loop is moving in its own plane with velocity $$v$$ in a uniform magnetic field perpendicular to its plane as shown in the figure. An electric field is induced

A
in AD, but not in BC

B
in BC, but not in AD

C
neither AD nor BC

D
in both AD and BC

Solution

Both AD and BC are straight conductors moving in a uniform magnetic field and emf will be induced in both. This will cause electric field in both.
Question 5
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A long solenoid has $$500\ turns$$. When a current of $$2\ A$$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $$4 \times {10}^{-3}\ Wb$$. The self-inductance of the solenoid is:

A
$$4.0\ H$$

B
$$2.5\ H$$

C
$$2.0\ H$$

D
$$1.0\ H$$

Solution

Given:-
$$\Phi=4\times 10^{-3}Wb/turn$$
$$N=$$ Number of turns $$=500$$
$$I=$$ Current flowing $$=2A$$
$$L=$$ Self-inductance $$=?$$
Formula:-
$$N\times \Phi=L\times I$$

$$\therefore 500\times 4\times 10^{-3}=L\times 2$$

$$\therefore L=\dfrac{500\times 4\times 10^{-3}}{2}$$

$$\therefore L=1H$$

$$\therefore$$ Self-inductance of solenoid is $$1.0H$$
Question 6
1 / -0

A solid conducting sphere of radius $$R$$ is moved with a velocity $$V$$ in a uniform magnetic field of strength $$B$$ such that $$\bar { B }$$ is perpendicular to $$\bar {V}$$. The maximum e.m.f. induced between two points of the sphere is :

A
$$RBV$$

B
$$\sqrt { 2 }\ RBV$$

C
$$\dfrac { RBV }{ 2 }$$

D
$$2\ RBV$$

Solution
Question 7
1 / -0

The armature of a generator of resistance $$1\Omega$$ is rotated at its rated speed and produces $$125V$$ without load and $$115V$$ with full load. The current in the armature coil is

A
$$240A$$

B
$$10A$$

C
$$1A$$

D
$$2A$$

Solution

$$R=1\Omega$$
$$E=125V$$
$$e=115V$$ (Back emf)

we know,

$$I=\dfrac{V}{R}$$

$$I=\dfrac{125-115}{1}$$

$$I=10 A$$

So, the current in armature coil is 10 ampere
Question 8
1 / -0

A circular coil of mean radius of 7 cm and having 4000 tums is rotated at the rate of 1800 revolutions per minute in the earth's magnetic field (B = 0.5 Gauss ) , The peak value of emf induced is

A
1 . 158 V

B
0 . 58 V

C
0 . 29 V

D
5 . 8 V

Solution

Given,

Angular frequency $$\omega =2\pi f=\dfrac{2\pi 1800}{60}=60\pi $$

Magnetic field, $$B=0.5\ G=0.5\times {{10}^{-4}}\ T$$

Area, $$A=\pi \times {{0.07}^{2}}=0.0154\,{{m}^{2}}$$

Flux $$\phi =nBA\cos \theta =nB(\pi {{r}^{2}})\cos \omega t$$

The induced emf is

$$\varepsilon =\dfrac{-d\phi }{dt}=nBA\omega \sin \omega t$$

For maximum E.M.F, $$\varepsilon ={{\varepsilon }_{o}},\ \sin \omega t=1$$

$$ {{\varepsilon }_{o}}=n\omega BA $$

$$ {{\varepsilon }_{o}}=4000\times 60\pi \times 0.5\times {{10}^{-4}}\times 0.0154 $$

$$ {{\varepsilon }_{o}}=0.58\,V $$
Question 9
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A rod of length $$50\ cm$$ moves with a speed of $$10\ cm/s$$, in a uniform magnetic field of strength $$10\ G$$ at an angle of $$30^{o}$$ with the field. The emf induced across the ends of the rod is :

A
$$5000\ CGS\ unit$$

B
$$2500\ CGS\ unit$$

C
$$7500\ CGS\ unit$$

D
$$1000\ CGS\ unit$$

Solution

Given,

Velocity, $$V=10\,cm/s$$

Length of rod, $$L=50\,cm$$

Magnetic Field, $$B=10\,G$$

Motional $$emf=BlV\sin \theta =10\times 50\times 10\sin {{30}^{o}}=2500\,\ CGS\ unit$$
Question 10
1 / -0

The magnetic flans associated with a metal ring varies with time all. to $$\phi = 3 (at^3 - bt^2)$$ $$Tm^2, a = 2sec^{-3}, b = 6sec^{-2}$$. If the resistance of the ring is $$24 \Omega$$, the current induced in the ring during the time $$t = 2$$ sec is

A
$$2 amp$$

B
$$4 amp$$

C
$$6 amp$$

D
$$zero$$

Solution

Given,

$$\phi=3(at^3-bt^2)Tm^2$$

$$a=2 sec^{-3}$$

$$b=6 sec^{-3}$$

$$R=24\Omega$$

The induced emf

$$e=IR=\dfrac{d\phi}{dt}=3(3at^2-2bt)$$

$$I=\dfrac{3}{R}(3at^2-2bt)$$

At $$t=2sec$$

$$I=\dfrac{3}{24}(3\times 2\times 4-2\times 6\times 2)$$

$$I=0$$ $$amp$$

The correct option is D.