Electromagnetic Induction Test

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Electromagnetic Induction Test - 71

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Question 1
1 / -0

An alternating current generator has an internal resistance $$R_{g}$$ and an internal reactance $$X_{g}$$. It is used to supply power to a passive load consisting of a resistance $$R_{g}$$ and a reactance $$X_{L}$$. For maximum power to be delivered from the generator to the load, the value of $$X_{L}$$ is equal to

A
Zero

B
$$X_{g}$$

C
$$-X_{g}$$

D
$$R_{g}$$

Solution

Generator
$$R_g , X_g$$

Passive load
$$R_g , X_L$$

The condition for the power to be maximum is the veactance should be zero.
$$X_g + X_L = 0$$

$$X_L = -X_g$$
Question 2
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A pair of parallel conducting rails lie at right angle to a uniform magnetic field of $$2.0\ T$$ as shown in the figure. Two resistors $$10\ \Omega$$ and $$5\ \Omega$$ are to slide without friction along the rail. The distance between the conducting rails is $$0.1\ m$$. Then:

A
$$=\dfrac {1}{150}$$ induced current $$A$$ directed anticlockwise if $$10\ \Omega$$ resistor is pulled to the right with speed $$0.5\ ms$$ and $$5\ \Omega$$ resistor is held fixed

B
induced current $$=\dfrac {1}{300}\ A$$ directed anticlockwise if $$10\ \Omega$$ resistor is pulled to the right with speed $$0.5$$ and $$5\ \Omega$$ resistor is held fixed

C
$$=\dfrac {1}{300}$$ induced current $$A$$ directed clockwise if $$5\ \Omega$$ resistor is pulled to the left at $$0.5\ {ms}^{-1}$$ and $$10\ \Omega$$ resistor is held fixed

D
$$=\dfrac {1}{150}$$ induced current $$A$$ directed clockwise if $$5\ \Omega$$ resistor is pulled to the left at $$0.5\ {ms}^{-1}$$ and $$10\ \Omega$$ resistor is held at rest.

Solution

We are given,
magnetic field, $$B=2T$$
resistance of AB = $$5\Omega$$
resistance of CD = $$10\Omega$$
distance b/w rails, $$l=0.1m$$
The emg induced (E) where one wire is moved with velocity $$(v)=0.5m/s$$ and other wire is held fixed.
$$\epsilon=Blv$$
$$=2\times 0.1\times 0.5$$
$$=\epsilon=0.1V$$
Now, since the resistance of wires AB and CD will be in series, So Req
$$Req=10+5$$
$$Req=15\Omega$$
Now, the induced current, I is
$$I=\dfrac{\epsilon}{Req}$$
$$=\dfrac{0.1}{15}$$
$$\boxed{I=\dfrac{1}{150}A}$$
And, by right hand rule, the direction of induced current us clockwise.
Thus, the induced current is $$\dfrac{1}{150}A$$ when a wire is moved with velocity $$0.5m/s$$. The direction of induced current is anticlockwise.
Question 3
1 / -0

The current through an inductor of $$1\ H$$ is given by
$$ i=3t\ \sin { t }$$
The voltage across the inductor of $$1\ H$$ is:

A
$$3\ \sin { t } +3\ \cos { t }$$

B
$$3\ \cos { t } +t\ \sin { t }$$

C
$$3\ \sin { t } +3t\ \cos { t }$$

D
$$3t\ \cos { t } +\ \sin { t }$$

Solution
Question 4
1 / -0

A magnetic field given by $$B(t) = 0.2 t - 0.05 { t }^{ 2 }$$ tesla is directed perpendicular to the plane of a circular coil containing $$25$$ turns of radius $$1.8\ cm$$ and whose total resistance is $$1.5\ \Omega $$. The power dissipation at $$3\ s$$ is approximate.

A
$$1.37 \mu W$$

B
$$7 \mu W$$

C
zero

D
$$4 \mu W$$

Solution
Question 5
1 / -0

The diagram below shows two coils A and B placed parallel to each other at a very small distance. Coil A is connected to an ac supply. G is a very sensitive galvanometer. When the key is closed -

A
Constant deflection will be observed in the galvanometer for 50 Hz supply

B
Visible small variations will be observed in the galvanometer for 50 Hz input

C
Oscillations in the galvanometer may be observed when the input ac voltage has a frequency of 1 to 2 Hz

D
No variation will be observed in the galvanometer even when the input ac voltage is 1 or 2 Hz

Solution

When the switch , K is closed there will be AC current in the loop A. Due to this AC, there will be induced current in the loop nearly to A which is B. So, Galvanometer (G) will sense the current.
When the frequency of oscillation, the galvanometer can be seen to have oscillations (deflection)
But when frequency is high, the oscillations in the galvanometer will so fast, that it may seem to not vary in its reading.
Thus, oscillation in galvanometer may be observed when input AC voltage has low frequency (1-2Hz)
Question 6
1 / -0

Two conducting rings P and Q of radii and r and 2r rotate uniformly in opposite directions with centre of mass velocities 2 v and v respectively on a conducting surface S. There is a uniform magnetic field of magnitude B perpendicular to the plane of the rings. The potential difference between the highest points of the two rings is

A
(a) zero

B
(b) 4 Bvr

C
(c) 8 Bvr

D
(d)16 Bvr

Solution

Effective length moving in magnetic field is $$2r$$ and $$4r$$ respectively.

$$\therefore E_1 = B(2r)(2v) = 4 Bvr$$

$$ E_r = B(4r)(v) = 4 Bvr$$

To find the potential difference, we replace the induced emfs in the rings by cells.

$$\therefore$$ Potential Difference $$= E_1 + E_2 = 8 Bvr$$
Question 7
1 / -0

Current $$1$$ in a long $$4y-$$axis is paced through a square metal frame of side $$2a$$ oriented in the $$y-z$$ p[lane a shown. The linear mass density of the frame is $$\lambda$$. A uniform magnetic field $$B$$ is now switched on along $$x-$$axis. Then the instantaneous angular acceleration of the frame will be:

A
$$\dfrac {4IB}{\lambda a}$$

B
$$\dfrac {12IB}{\lambda a}$$

C
$$\dfrac {4IB}{3 \lambda a}$$

D
$$zero$$

Solution

Linear mass density is $$\lambda$$
Current I is along $$4y-axis$$ through square metal frame of side $$2a$$
Instantaneous acceleration :- $$\dfrac{IB}{\lambda a}=\dfrac{4IB}{\lambda q}$$
Question 8
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In figure shown, wire $$P_{1}Q_{1}$$ and $$P_{2}Q_{2}$$, both are moving towards right with speed $$5\ cm/sec$$. Resistance of each wire is $$2\ \Omega$$ . Then current through $$19\ \Omega $$ resistor is:

A
$$0$$

B
$$0.1\ mA$$

C
$$0.2\ mA$$

D
$$0.3\ mA$$

Solution
Question 9
1 / -0

When the current in a coil changes from 8 ampere to 2 ampere in $$3 \times 10^{-2}$$ second, the e.m.f. induced in the coil is 2 volt. The self inductance of the coil (in millinery) is

A
1

B
5

C
20

D
10

Solution

$$E.M.F. = L \dfrac{di}{dt}$$

$$2 = L \times \dfrac{8-2}{3 \times 10^{-2}}$$

L = 1 millinery

Here (A) is correct answer
Question 10
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A circular wire $$ABC$$ and a straight conductor $$ADC$$ are carrying current $$i$$ and are kept or the magnetic field $$B$$ then considering points $$A$$ and $$C$$

A
Force on $$ABC$$ is more than that on $$ADC$$

B
Force on $$ABC$$ is less than that on $$ADC$$

C
Force on $$ABC$$ is equal to that on $$ADC$$

D
any of $$(A)$$ or $$(B)$$ or $$(C)$$

Solution

Force a conductor placed in a magnetic field is given by

$$F = BIl\sin \theta $$

Here for both conductors $$\theta = 90^\circ $$, so $$\sin \theta = 1$$.

The length of the conductor ABC is more than conductor ADC.

So force on conductor ABC will be more than force on conductor ADC.