Electromagnetic Induction Test

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Electromagnetic Induction Test - 74

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Question 1
1 / -0

A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction B. At the position MNQ, the speed of the ring is v. The potential difference developed across the ring is

A
zero

B
$$1/2 \,B \,v \pi R^2$$ , and M is at a higher potential

C
$$\pi\,R\,Bv$$, and Q is at a higher potential

D
$$2\,R\,Bv$$, and Q at a higher potential

Solution
Question 2
1 / -0

Two circuits, each consisting of a loop of wire are placed in the proximity of one another and one of them is carrying a current. The current is then suddenly stopped. Then

A
There will be an induced e.m.f. in the other circuit

B
There will be no induced e.m.f

C
There will be an induced e.m.f. in both the circuits

D
None of these

Solution
Question 3
1 / -0

A coil having $$n$$ turns and resistance $$4R$$ $$ \Omega$$. This combination is moved in time $$t$$ seconds from a magnetic field $$W_1$$ weber to $$W_2$$ weber. The induced current in the circuit is :

A
$$-\dfrac{W_2-W_1}{5Rnt}$$

B
$$-\dfrac{(W_2-W_1)}{5Rnt}$$

C
$$\dfrac{W_2-W_1}{Rnt}$$

D
$$\dfrac{n(W_2-W_1)}{5Rt}$$
Question 4
1 / -0

A train is moving from south to north with a velocity of $$90$$ km/h. The vertical component of earth's magnetic induction is $$0.4\times { 10 }^{ -4 }\quad Wb/{ m }^{ 2 }.$$ If the distance between the two rails is $$1 m,$$ what is the induced e.m.f. in its axle?

A
$$1mV$$

B
$$0.1mV$$

C
$$10mV$$

D
$$100mV$$

Solution

By taraday's law,
Einduced - $$ \frac{-d\phi }{dt}$$
$$ \phi = \vec{B}.\vec{A} = |\vec{B}| |\vec{A}| cos\theta $$
Here $$ \theta = 0$$ so,
$$ \phi = B.A$$
$$ \dfrac{-d\phi }{dt} = \dfrac{dB.A}{dt} = \dfrac{-BdA}{dt}-\dfrac{AdB}{dt}$$
As B is constant, $$ \dfrac{dB}{dt} = 0$$
$$ = \dfrac{d\phi }{dt} = \dfrac{-B.dA}{dt}$$
Here, $$ \dfrac{dA}{dt} = \dfrac{dlb}{dt} = \dfrac{ldb}{dt}+\dfrac{bdl}{dt}$$
b(breadth) of the train is not
changing it a length that is going
into the field is varying with speed
90km/h(25m/s),so.
$$ \dfrac{dA}{dt} = l(0)+bv = (1)(25) = 25$$
so,$$ \dfrac{d\phi }{dt} = -(0.4) \times 10^{-4} \times 25 = 10^{-3}v$$
Einduced =$$ 10^{-3}v = 1 mv$$
Option - A is correct.
Question 5
1 / -0

In a given figure an isosceles triangle conducting wire is entering into the region of perpendicular uniform magnetic field $$B_0$$ with constant velocity $$V_0$$. The magnitude of induced emf as a function of time(t) in the wire is(till side is outside the field)

A
$$B_0V^2_0t^2$$

B
$$2B_0V^2_0t^2$$

C
$$2B_0V^2_0t$$

D
$$2B_0V_0t$$

Solution
Question 6
1 / -0

A magnet is brought near a coil in two ways (i) rapidly (ii) slowly. The induced charge will be

A
More in case (i)

B
More in case (ii)

C
Equal in both the cases

D
More or less according to the radius of the coil

Solution
Question 7
1 / -0

A wire of fixed length is wound on a solenoid of length l and radius r. Its self inductance is found to be L.Now if same wire is wound on a solenoid of length $$\dfrac{l}{1}$$ and radius $$\dfrac{r}{2}$$, then self inductance will be

A
2L

B
$$\dfrac{L}{2}$$

C
3L

D
4L

Solution
Question 8
1 / -0

The coefficient of mutual inductance of two coils is $$6mH$$. If the constant current of $$2A$$ is flowing in one coil , then the induced e.m.f in the second coil will be

A
$$3mV$$

B
$$2mV$$

C
$$3V$$

D
Zero

Solution
Question 9
1 / -0

An electron originates at a point $$A$$ lying on the axis of a straight solenoid and moves with velocity $$v$$ at an angle $$\alpha$$ to the axis. The magnetic induction of the field is equal to $$BA$$ screen is oriented at right angles to the axis and is located at a distance $$1$$ from the point $$a$$. Find the distance from the axis to the point on the screen into which the electron strikes.

A
$$d = 5r\sin \left (\dfrac {\theta}{2}\right )$$, Here $$r = 2\dfrac {mv\sin \alpha}{eB}$$ and $$\theta = \dfrac {eBl}{mv\cos \alpha}$$.

B
$$d = 2r\sin \left (\dfrac {\theta}{2}\right )$$, Here $$r = \dfrac {mv\sin \alpha}{eB}$$ and $$\theta = \dfrac {eBl}{mv\cos \alpha}$$.

C
$$d = 3r\sin \left (\dfrac {\theta}{2}\right )$$, Here $$r = 3\dfrac {mv\sin \alpha}{eB}$$ and $$\theta = \dfrac {eBl}{mv\cos \alpha}$$.

D
$$d = 4r\sin \left (\dfrac {\theta}{2}\right )$$, Here $$r = \dfrac {mv\sin \alpha}{eB}$$ and $$\theta = \dfrac {eBl}{mv\cos \alpha}$$.
Question 10
1 / -0

An $$L$$-shaped conductor rod is moving in transverse magnetic field as shown in the figure. Potential difference between ends of the rod is maximum if the rod is moving with velocity :

A
$$3\hat{i} - 6\hat{j} \ m/s$$

B
$$-4\hat{i} - 6\hat{j} \ m/s$$

C
$$3\hat{i} - 2\hat{j} \ m/s$$

D
$$\sqrt{13}\hat{i} \ m/s$$

Solution