Electromagnetic Induction Test

Result

Electromagnetic Induction Test - 80

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Three resistance of magnitude $$R$$ each are connected in the form of an equilateral triangle of side $$a$$. The combination is placed in a magnetic field $$B=B_{0}e^{-\lambda t}$$
perpendicular to the plane. The induced current in the circuit is given by:

A
$$\left(\dfrac{a^{2}\lambda}{2\sqrt{3R}}B_0\right)e^{-\lambda t}$$

B
$$\left(\dfrac{a^{2}\lambda}{4(\sqrt{3})R}B_0\right)e^{-\lambda t}$$

C
$$\left(\dfrac{a^{2}B_0}{\lambda 4\sqrt{3R}}\right)e^{-\lambda t}$$

D
$$\left(\dfrac{a^{2}B_0 R}{\lambda 4\sqrt{3R}}\right)e^{-\lambda t}$$

Solution
Question 2
1 / -0

Consider the situation shown in figure. If the closed loop is completely enclosed in the circuit containing the switch, the closed loop will show____

A
an anticlockwise current-pulse

B
a clockwise current-puls

C
an anti-clockwise current-pulse and then a clockwise current-pulse

D
a clockwise current-pulse and then an anticlockwise current-pulse

Solution
Question 3
1 / -0

A wire is sliding as shown in the figure. The angle between the acceleration and velocity of the wire is

A
$$30^{o}$$

B
$$40^{o}$$

C
$$120^{o}$$

D
$$90^{o}$$

Solution
Question 4
1 / -0

A solenoid of inductance $$L$$ and resistance $$r$$ is connected in parallel to a resistance $$R$$ and a battery of emf $$E$$. Initially if the switch is closed for a long time and at $$t=0$$, then the :

A
current through solenoid at any time $$t$$, after opening the switch is $$\dfrac{E}{r}e^{-(R+r)t/L}$$

B
induced emf across solenoid at time $$t=0$$ is $$\dfrac{E(R+r)}{r}$$

C
amount of heat generated in solenoid is $$\dfrac{E^2L}{2r(r+R)}$$

D
potential difference across solenoid at $$t=0$$ is $$E$$

Solution
Question 5
1 / -0

An Indian ship with a vertical conducting mass navigates the Indian ocean in the latitude of magnetic equator. To induce the greatest emf in the mast, the ship should proceed:

A
northward

B
southward

C
eastward

D
none of these

Solution
Question 6
1 / -0

A short solenoid (length $$l$$ and radius $$r$$, with $$n$$ turns per unit length) lies well inside and on the axis of a very long, coaxial solenoid (length $$L$$, radius $$R$$ and $$N$$ turns per unit length, with $$R > r$$). Current $$I$$ flows in the short solenoid.
Choose the correct statement.

A
There is uniform magnetic field $$\mu_0n I$$ in the long solenoid.

B
Mutual inductance of the solenoid is $$\pi \mu_0 r^2 n Nl$$.

C
Flux through outer solenoid due to current $$I$$ in the inner solenoid is proportional to the ratio $$R/r$$

D
Mutual inductance of the solenoids is $$\pi \mu_0 r RnNl L/(rR)^{1/2}$$

Solution

The magnetic flux linked with long solenoid of radius $$R$$ due to short solenoid of radius $$r$$ having current $$I$$ is,

$$\phi=MI$$. . . . .(1)

The magnetic field due to the short solenoid,

$$B=\mu_0 nl I$$

Magnetic flux linked is,

$$\phi= BA$$

$$\phi= \mu_0nlI(\pi r^2)$$. . . . .(2)

From equation (1) and (2), we get,

$$M=\pi \mu_0r^2nl$$

For $$N$$ number of turns,

$$M=\pi \mu_0r^2nNl$$.

The correct option is B.
Question 7
1 / -0

In fig a square loop PQRS of side a and resistance r is placed neat an infinitely long wire carrying a constant current l. the sides PQ and Rs parallel to the wire. The wore and the loop are in the same plane. The loop is rotated $$180^{\circ}$$ about an axis parallel to the long wire and passing through the mid-points of the sides QR and PS.The total amount of charge which passes through any point of the loop during rotation is

A
$$\dfrac {\mu_0Ia}{2 \pi r}ln2$$

B
$$\dfrac {\mu_0Ia}{\pi r}ln2$$

C
$$\dfrac {\mu_0Ia^2}{2 \pi r}$$

D
Cannot be found because time of rotation is not give.

Solution

As we know,

Magnetic flux density at a distance x from the infinitely long current-carrying wire is,
$$B=\dfrac{\mu I }{2\pi x}$$

Initial flux: $$\phi_i=- \int_{a}^{2a} \dfrac {\mu_0I_a}{2\pi} \dfrac {dx}{x}$$

$$=-\dfrac {\mu_0Ia}{2 \pi }ln2$$

Final flux :$$\phi_f=\dfrac {\mu_0Ia}{2 \pi }ln2$$

Charge flown:$$q=\dfrac {\Delta \phi}{r}=\dfrac {\phi_f-\phi_i}{r}=\dfrac {\mu_0Ia}{\pi r}ln2$$
Question 8
1 / -0

A metal bar is moving with a velocity of $$v=5 cm s^{-1}$$ over a U-shaped conductor.AT $$t=0$$, the external magnetic field is 0.1 T directed out of the page and is increasing at a rate of $$0.2 T s^{-1}$$. Take $$l= 5 cm$$ and $$ t=0, x=5 cm$$

The current flowing in the circuit is

A
2.5 A

B
5 A

C
1 A

D
2 A

Solution

Let us take anticlockwise direction as positive, then the area vector in the upward direction will be positive.
now as we know the flux is
$$\phi= BA=Blx$$

$$e= -{d\phi}{dt}=-l [B \dfrac {dx}{dt}+x \dfrac {dB}{dt}]$$

$$=-5 \times 10^{-2}[ 0.1 \times (-5 \times 10^{-2} + 5 \times 10^{-2} \times 0.2]$$

$$-250 \times 10^{-6} V= -250 \mu V$$

emf is coming out to be negative, so it should be clockwise.

Current $$I= \dfrac {e}{R}= \dfrac {250 \times 10^{-6}}{10^{-4}}=2.5A$$
Question 9
1 / -0

An aircraft is flying at a level height in presence the magnetic field of the Earth. If an electric bulb is connected between the two extreme ends of the wings,

A
a voltage will induce across the wings and the bulb will glow.

B
no voltage will induce across the wings but the bulb will glow.

C
a voltage will induce across the wings but the bulb will not glow.

D
no voltage will induce across the wings and the bulb will not glow.

Solution
Question 10
1 / -0

The radius of the circular conducting loop shown in fig. is R. Magnetic field is decreasing at a constant rate $$\alpha$$ Resistance per unit length of the loop is $$\rho$$then the current in wire AB is (AB is one of the diameter)

A
$$\dfrac {R \alpha}{2\rho}$$ from A to B

B
$$\dfrac {R \alpha}{2\rho}$$ from B to A

C
$$\dfrac {R \alpha}{\rho}$$ from A to B

D
0

Solution