Electromagnetic Induction Test

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Electromagnetic Induction Test - 70

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Weekly Quiz Competition

Question 1
1 / -0

A wheel with 20 metallic spokes each of length 0.8 m long is rotated with a speed of 120 revolution per minute in a plane normal to the horizontal component of earth magnetic field H at a place. If H=$$0.4\times 10^{-4}$$ T at the place, then induced emf between the axle and the rim of the wheel is

A
$$2.3\times 10^{-4}$$

B
$$3.1\times 10^{-4}$$

C
$$2.9\times 10^{-4}$$

D
$$1.61\times 10^{-4}$$

Solution

Here, $$H = B = 0.4 \times 10^{-4} T, l = 0.8 m \,\,\, v = 120 rpm = 2 rps $$

Emf induced across the ends of each spoke
$$\epsilon = \dfrac{1}{2} B \omega l^2 = \dfrac{1}{2} B(2 \pi \upsilon)l^2$$ $$\,\,\,\,\,\,\,$$ $$(\because \omega = 2 \pi \upsilon)$$

$$= \, B \pi \upsilon l^2$$
$$ \therefore \epsilon = 0.4 \times 10^{-4} \times \pi \times 2 \times (0.8)^2 = 1.61 \times 10^{-4} \, V$$

Note : Number of spokes is not relevant because the emfs across the spoke area is parallel.
Question 2
1 / -0

When a coil rotated in magnetic field induced current in it :

A
continuously changes

B
remains same

C
becomes zero

D
becomes maximum

Solution

We know that EMF induced=$$E=NBA\omega\cos\omega t$$

$$\implies I=\dfrac{E}{R}=\dfrac{NBA\omega}{R}\cos\omega t$$

Hence, current continuously changes with time.

Answer-(A)
Question 3
1 / -0

As shown in figure, two vertical conducting rails separated by distance $$1.0\ m$$ are placed parallel to z-axis. At $$z = 0$$, a capacitor of $$0.15\ F$$ is connected between the rails and a metal rod of mass $$100\ gm$$ placed across the rails slides down along the rails. If a constant magnetic fields of $$2.0\ T$$ exists perpendicular to the plane of the rails, what is the acceleration of the rod?
(Take $$g = 9.8\ m/s^{2})$$.

A
$$2.5\ m/s^{2}$$

B
$$1.4\ m/s^{2}$$

C
$$9.8\ m/s^{2}$$

D
$$0$$

Solution

Due to motion of rod, emf induced across capacitor, $$\epsilon = Blv$$
$$\therefore$$ Charge stored in capacitor, $$Q = C(Blv)$$
$$I = \dfrac {dQ}{dt} = CBl \dfrac {dv}{dt} = CBla$$
Force opposing the downward motion, $$F_{m} = BIl$$
$$\therefore F_{m} = B(CBla)l = B^{2}l^{2}Ca$$
Net force on rod, $$F_{net} = W - F_{m} = mg - B^{2}l^{2}CA$$
$$\therefore ma = mg - B^{2}l^{2}Ca$$
or $$a = \dfrac {mg}{(m + B^{2}l^{2}C)}$$
So, $$a = \dfrac {0.1\times 9.8}{(0.1 + 2^{2} \times 1^{2}\times 0.15)} = 1.4\ m/s^{2}$$.
Question 4
1 / -0

An air-cored solenoid with length $$20 cm$$, area of cross-section is $$20 cm^2$$ and the number of turns $$400$$ carries a current $$2 A$$. The current is suddenly switched off within $$10^{-3}$$ sec. The average back emf induced across the end of the open switch in the circuit is ( ignore variation in the magnetic field near the end of the solenoid)

A
$$2 V$$

B
$$4 V$$

C
$$3 V$$

D
$$5 V$$

Solution

Given:
$$ l= 20 cm= 20 \times 10^{-2}m $$
$$A=20cm^2=20\times 10^{-4} m^2$$
$$N=400, $$
$$I_1= 2A,I_2=0, dt=10^{-3}s$$

$$\epsilon= \dfrac {d\phi}{dt}=\dfrac{d(BAN)}{dt}$$

$$=\dfrac{\mu_0 NdIAN}{ldt}$$

$$=\dfrac{\mu_0 N(I_1-I_2)AN}{ldt}$$

$$= \dfrac {d\phi}{dt}=\dfrac{d(BAN)} {dt}$$

$$\epsilon=\dfrac{4\pi\times\,10^{-7}\times(400)^2\times 2\times\times 10^{-4}}{20\times 10^{-2}\times 10^{-3}}$$

$$\epsilon=4$$V
Question 5
1 / -0

A current $$i$$ is flowing in a straight conductor of length $$L$$. The magnetic induction at the point on its perpendicular bisector at a distance $$L/4$$ from its centre will be

A
zero

B
$$\dfrac{\mu_{0}i}{\sqrt{2}\pi L}$$

C
$$\dfrac{\mu_{0}i}{\sqrt{2}L}$$

D
$$\dfrac{4\mu_{0}i}{\sqrt{5}\pi L}$$

Solution
Question 6
1 / -0

Figure shows a circular area of radius $$R$$ where a uniform magnetic field $$\vec { B }$$ is going into the plane of paper and increasing in magnitude at a constant rate. In rate case, which of the following graphs, drawn schematically, correctly shown the variation of the induced electric field $$E(r)$$?

A

B

C

D
Question 7
1 / -0

A bulb of rated values 60 V and 10 W is connected in series with a source of 100 V and 50 Hz. The coefficient of self induction of a coil to be connected in series for its operation will be:

A
1.53 H

B
2.15 H

C
3.27 H

D
3.89 H
Question 8
1 / -0

One of the two small circular coils, (none of them having any self-inductance) is suspended with a $$V-$$shaped copper wire, with plane horizontal. The other coil is placed just below the first one with plane horizontal. Both the coil are connected in series with a $$dc$$ apply. The coils are found to attract each other with a force. Which one of the following statement is incorrect?

A
Both the coils carry currents in the same direction.

B
Coils will attract each other, even if the supply is an $$ac$$ source.

C
Force is proportional to $${d}^{-1}, d=$$ distance between the centers of the coil.

D
Force is proportional to $${d}^{-2}$$.

Solution
Question 9
1 / -0

Two coils, $$X$$ and $$Y$$, are kept inclose vicinity of each other. When a varying current, $$l(t)$$, flows through coil $$X$$, the induced emf $$(V(t))$$ in coil $$YY$$, varies in the manner shown here. The variation of $$l(t)$$, with time, can then represented by the graph labelled as graph.

A

B

C

D

Solution
Question 10
1 / -0

A coil of inductance $$8.4\ mH$$ and resistance $$ 6 \Omega$$ is connected to a $$12\ V$$ battery. The current in the coil is $$1.0\ A$$ at the time, approximately

A
$$500\ s$$

B
$$20\ s$$

C
$$35\ ms$$

D
$$1\ ms$$

Solution

In the given growth circuit as,
$$i=\dfrac{V}{R}\left(1-e^{-\dfrac{Rt}{2}}\right)$$
Here, $$V=12v$$
$$R=6\Omega$$ and $$L=8.4\ mH$$
$$\Rightarrow i=2\left(1-e^{\dfrac{-10^3t}{1.4}}\right)$$
For $$i=1\ A$$,
$$1=2\left(1-e^{\dfrac{-10^3t}{1.4}}\right)$$
$$1/2=1-e^{\dfrac{10^3t}{1.4}}$$
$$\Rightarrow e^{\dfrac{-10^3t}{1.4}}=\dfrac{1}{2}\rightarrow t=\dfrac{1.4(\ln 2)}{10^3}s$$
$$\Rightarrow t=(1.4)(0.69)\ ms=1\ ms$$
option $$-D$$ is correct.