Electromagnetic Induction Test

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Electromagnetic Induction Test - 50

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Question 1
1 / -0

A rod of length $$l$$ rotates with a uniform angular velocity $$\omega$$ rad/s about an axis passing through its middle point but normal to its length in a uniform magnetic field of induction $$B$$ with its direction parallel to the axis of rotation. The induced emf between the two ends of the rod is :

A
$$\dfrac{Bl^2\omega}{2}$$

B
$$zero$$

C
$$\dfrac{Bl^2\omega}{4}$$

D
$$2Bl^2\omega$$

Solution

Note that an EMF always generated along the direction $$(\vec V \times \vec B)$$
For circular motion, $$\vec V=\vec R \omega$$

Hence, EMF induced, $$\varepsilon= \omega (\vec R \times \vec B)$$

Now, the two halves of the rod equal, $$OA=OB$$
thus, $$|\vec{R_1}|=|\vec{R_2}|$$

but $$\vec{R_1}=-\vec{R_2}$$.

Therefore, the induced EMFs will be in the opposite direction from the point $$O$$.

Thus two EMFs $$\varepsilon_1$$ and $$\varepsilon_2$$ are generated in equal and opposite direction, leading to a net ZERO potential difference across
the ends.
Question 2
1 / -0

The total charge induced in a conduction loop when it is moved in magnetic field depends on

A
The rate of change of magnetic flux

B
Initial magnetic flux only

C
The total change in magnetic flux

D
Final magnetic flux only

Solution
Question 3
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A proton of mass 'm' moving with a speed v ($$< <$$c, velocity of light in vacuum) completes a circular orbit in time 'T' in a uniform magnetic field. If the speed of the proton is increased to $$\sqrt{2}$$v, what will be time needed to complete the circular orbit?

A
$$\sqrt{2}T$$

B
T

C
$$\displaystyle\frac{T}{\sqrt{2}}$$

D
$$\displaystyle\frac{T}{2}$$

Solution

$$T=\displaystyle\frac{2\pi m}{qB}$$
T is independent of v.
Question 4
1 / -0

A jet plane is travelling towards west at a speed of $$1800\ km/h$$. What is the voltage difference developed between the ends of the wing having a span of $$25\ m$$, if the Earth's magnetic field at the location has a magnitude of $$5 \times {10}^{-4}\ T$$ and the dip angle is $$30^o$$.

A
0.31 V

B
3.1 V

C
5 V

D
10 V

Solution

$$v=1800\quad km/h=1800\times \dfrac { 5 }{ 18 } =500m{ s }^{ -1 }$$
$$l\rightarrow $$ distance between the ends of the wings=$$25m$$
$$B=5\times { 10 }^{ -4 }T$$
$$\\ \theta =30°$$
Motional emf $$\varepsilon ={ B }_{ v }vl\\ $$ [$$B_v$$-verticle component of earth's magnetic field]
$$\varepsilon =5\times { 10 }^{ -4 }\times \sin { 30°\times 500\times 25=3.1V } $$
Question 5
1 / -0

There is a uniform magnetic field directed perpendicular and into the plane of the paper. An irregular shaped conducting loop is slowly changing into a circular loop in the plane of the paper. Then

A
current is induced in the loop in the anti-clockwise direction

B
current is induced in the loop in the clockwise direction.

C
ac is induced in the loop.

D
no current is induced in the loop

Solution

Due to change in the shape of the loop, the magnetic flux linked with the loop increases. Hence, current is induced in the loop in such a direction that it
opposes the increases in flux. Therefore, induced current flows in the anticlockwise direction.
Question 6
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Consider the situation shown in figure. The wire AB slides on the fixed rails with a constant velocity. If the wire AB is replaced by a semicircular wire, the magnitude of the induced current will

A
Increase

B
Remain the same

C
Decrease

D
Increase or decrease depending on whether the semicircle bulges towards the resistance or away from it.

Solution

In both cases,$$e=V\times B\times dl=VBl$$
(option B is correct)
Question 7
1 / -0

A circular copper disc 10 cm in diameter rotates at 1800 revolution per minute about an axis through its centre and at right angles to disc. A uniform field of induction B of 1 Wb $$m^2$$ is perpendicular to disc. What potential difference is developed between the axis of the disc and the rim ?

A
0.023 V

B
0.23 V

C
23 V

D
230 V

Solution

Here,
$$l = r = 5\, cm = 5 \times 10^{-2} m,$$
B = 1 Wb $$m^{-2}$$

$$ \omega = 2 \pi \left( \dfrac{1800}{60} \right) \, rad \, s^{-1} = 60 \pi \, rad \, s^{-1},$$

$$\epsilon \, = \, \dfrac{1}{2} Bl^2 \omega \, =\, \dfrac{1}{2} \times 1 \times (5 \times 10^{-2})^2 \times 60 \pi = 0.23 V$$
Question 8
1 / -0

The mutual inductance $$M_{12}$$ of a coil 1 with respect to coil 2

A
increases when they are brought nearer

B
depends on the current passing through the coils.

C
increases when one of them is rotated about an axis.

D
both (a) and (b) are correct

Solution

Mutual Induction: Whenever the current passing through a coil or circuit changes, the magnetic flux linked with a neighbouring coil or circuit will also change. Hence an emf will be induced in the neighbouring coil or circuit. This phenomenon is called ‘mutual induction’.

If the two coils $$1$$ and $$2$$ are present with mutual inductance $$M_1$$ and $$M_2$$. Then the mutual inductance of the coil 1 due to 2 increases when they are bought near since, mutual inductance is proportional to the flux passed through the coil.

The mutual induction of $$M_{12}$$ is same as $$M_{21}$$
Question 9
1 / -0

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

A
maximum in situation (i)

B
maximum in situation (ii)

C
maximum in situation (iii)

D
same in all situation
Solution
Mutual induction $$∝$$ flux linkage between coils
Flux linkage in coil $$(i)$$ is maximum because both coils have same axis.
Flux linkage between coils is 0 because thus axis are perpendicular to each other.
Flux linkage between coils is 0 because their axis are perpendicular to each other .
So, mutual inductance between coils is maximum in case $$(i)$$.
Question 10
1 / -0

Two identical coaxial coils P and Q carrying equal amount of current in the same direction are brought nearer. The current in:

A
P increases while in Q decreases

B
Q increases while In P decreases

C
both P and Q increases

D
both P and Q decreases

Solution

Hint:- The direction of induced current is such that it opposes the cause by which it is generated. This is Lenz Law.

Explanation for the correct answer:
$$\bullet$$ It is given that there are two identical coaxial coils carrying current in equal amount in the same direction. This current carrying coil produces Magnetic Field around the coil, direction of which can be given by Right Hand Thumb Rule.
$$\bullet$$ When one coil carrying current is brought near another coil, the Magnetic Flux linked with the second coil changes, and according to Faraday's Law current is induced in the coil. According to Lenz Law the direction of induced current is opposite to that in the first coil.
$$\bullet$$ In the given situation, if P i brought near the coil Q, the current is induced in the coil Q in opposite direction. Thus, net current in coil Q decreases. Similarly net current in P also decreases.

Thus, net current in both P and Q decreases.
Option D is correct.