Electromagnetic Induction Test

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Electromagnetic Induction Test - 60

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Question 1
1 / -0

an equilateral triangular loop ADC having some resistance is pulled with a constant velocity v out of a uniform magnetic field directed into the paper. At time t=0, side DC of the loop is at edge of the magnetic field.The induced current I(i) versus time (t) graph will be

A

B

C

D

Solution

Let 2a be the side of the triangle and b be the length AE.

$$\dfrac {AH}{AE}=\dfrac {GH}{EC}\Rightarrow GH=(\dfrac {AH}{AE})EC$$

or $$GH=\dfrac {(b-vt)}{b}a=a-(\dfrac {a}{b}vt)$$

$$\therefore FG=2GH =2[a-\dfrac {a}{b}vt]$$

$$\therefore$$ Induced emf, , $$e=BV(FG)=2Bv(a-\dfrac {a}{b}vt)$$

$$\therefore$$ induced current, $$i=\dfrac {e}{R}=\dfrac {2Bv}{R}[a-\dfrac {a}{b}vt]$$

or $$i=K_1-K_2 t$$
Thus i-t graph is a straight line with a negative slope and positive intercept.
Question 2
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A vertical ring of radius r and resistance R falls vertically. it is in contact with two vertical rails which are joined at the top. The rails are without friction and resistance. There is a horizontal uniform magnetic field of magnitude B perpendicular to the plane of the ring and the rails. When the speed of the ring is v, the current in the top horizontal of the rail section is

A
0

B
$$\dfrac {2Brv}{R}$$

C
$$\dfrac {4Brv}{R}$$

D
$$\dfrac {8Brv}{R}$$

Solution

From the above figure,
The upper part of the ring has resistance R/2 and the lower part has resistance R/2.

$$I= \dfrac {2Brv}{\dfrac {R}{2}}=\dfrac {4Brv}{R}$$

Current in the top horizontal=$$2l=\dfrac{8Brv}{R}$$
Question 3
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A mutual inductor consists of two coils X and Y as shown in figure in which one quarter of the magnetic flux produced by X links with Y, giving a mutual inductance 'M'.
What will be the mutual inductance when Y is used as the primary?

A
M/4

B
M/2

C
M

D
2M

Solution

The mutual inductance 'M' remains the same whether X or Y is used as the primary. Mutual Inductance always depends upon the current flowing on other coil. M is directly proportional to flux linked from one coil to other coil. But flux is directly propotional to the current flowing through the coil.
Question 4
1 / -0

The peak value of an alternating emf E given by
$$ E = E_0 \,cos\, \omega t $$
is $$ 10\,V $$ and frequency is $$ 50\,Hz $$ . At time $$ t = (1/600) \,s $$ , the instantaneous value of emf is

A
$$ 10\,V $$

B
$$ 5\sqrt{3} \,V $$

C
$$ 5\,V $$

D
$$ 1\,V $$

Solution

Given , $$ E_0 = 10 \,V , t = \dfrac{1}{600},\,s $$

$$ \therefore $$ $$ E = E_0\,cos \,2\pi ft $$

$$ = 10\,cos\,\left [ 2\pi \times 50 \times \dfrac{1}{600} \right ] $$

$$ = 10 \, cos\,(\pi / 6) = 10\,(\sqrt{3} /2) = 5\sqrt{3} \,V $$
Question 5
1 / -0

Current in a coil of self-inductance 2.0 H is increasing as $$i=2\,sin\,t^2$$ . The amount of energy spent during the period when the current changes from 0 to 2 A is

A
1 J

B
2 J

C
3 J

D
4 J

Solution

Energy spent=$$\dfrac{1}{2}Ll^2$$
di= 2-0 = 2A
E $$ \dfrac{Ldi}{dt}$$
dw= $$\dfrac{Ldi}{dt}$$ i X dt
dw=Li di
Integrating the above solution from 0 to 2 A we get
W= $$ 2 \times( \dfrac{4}{2}-0) $$
W= 4 J
Question 6
1 / -0

A uniform magnetic field of induction B is confined to a cylindrical region of radius R. The magnetic field is increasing at a constant rate of $$dB/dt Ts^{-1}$$. An electron placed at the point P on the periphery of the field, experiences an acceleration

A
$$\dfrac {1}{2} \dfrac {eR}{m}{dB}{dt}$$ towards left

B
$$\dfrac {1}{2} \dfrac {eR}{m}{dB}{dt}$$ towards right

C
$$ \dfrac {eR}{m}{dB}{dt}$$ towards left

D
zero

Solution

At P induced electric field,

$$\int E.dl=-A\dfrac{dB}{dt}$$

$$E(2\pi R)=-(\pi R^2) \dfrac{dB}{dt}$$

The induced electric field at point P:

$$E= \dfrac {R}{2}\dfrac {dB}{dt}$$ towards right

acceleration of electron: $$a= \dfrac {eE}{m}=\ \dfrac {eR}{2m}\dfrac {dB}{dt}$$ towards left.
Question 7
1 / -0

A conductor AB of length l moves in xy plane with velocity $$\overrightarrow {v}=v_0 (\hat {i} -\hat {j})$$ A magnetic field $$\overrightarrow {B}=B_0 (\hat {i} +\hat {j})$$ exists in the region. The induced emf is

A
zero

B
$$2B_0lv_0$$

C
$$B_0lv_0$$

D
$$\sqrt 2B_0lv_0$$

Solution

As we know ,

Induced emf$$=l(\vec v\times \vec b)$$

Now from the above value of $$\vec v \ and\ \vec B$$

$$(\vec v\times \vec b)=0$$

hence , the induced emf =0.
Question 8
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The phenomenon of electromagnetic induction is :

A
the process of charging a body

B
the process of generating a magnetic field dur to current passing through a coil.

C
producing induced current in a coil by relative motion between a magnet and a coil.

D
the process of rotating a coil of an electric motor.

Solution

When there is relative motion between a magnet and a closed loop there is formation of induced emf due to change in magnetic flux linked with the coil. More the speed more the induced emf.
Question 9
1 / -0

The magnetic induction at the center $$O$$ in the figure shown is

A
$$\dfrac {\mu_0i}{4}\left( \dfrac {1}{R_1}-\dfrac {1}{R_2}\right)$$

B
$$\dfrac {\mu_0i}{4}\left( \dfrac {1}{R_1}+\dfrac {1}{R_2}\right)$$

C
$$\dfrac {\mu_0i}{4}\left( R_1-R_2\right)$$

D
$$\dfrac {\mu_0i}{4}\left( R_1+R_2\right)$$

Solution

In the following figure, magnetic fields at $$O$$ due to sections $$1, 2, 3$$ and $$4$$ are considered as $$B_1, B_2, B_3$$ and $$B_4$$ respectively.
$$B_1=B_3=0$$

$$B_2=\dfrac{\mu_0}{4\pi}.\dfrac{\pi i}{R_1}\otimes $$

$$B_4=\dfrac{\mu_0}{4\pi}.\dfrac{\pi i}{R_2}\odot$$

So $$B_{net}=B_2-B_4\Rightarrow B_{net}=\dfrac{\mu_0i}{4}\left( \dfrac{1}{R_1}-\dfrac{1}{R_2}\right) \otimes $$
Question 10
1 / -0

The magnetic induction due to an infinitely long straight wire carrying $$i$$ at a distance $$r$$ from wire is given by

A
$$|B|=\left( \dfrac {\mu_0}{4\pi}\right)\dfrac {2i}{r}$$

B
$$|B|=\left( \dfrac {\mu_0}{4\pi}\right)\dfrac {r}{2i}$$

C
$$|B|=\left( \dfrac {4\pi}{\mu_0}\right)\dfrac {2i}{r}$$

D
$$|B|=\left( \dfrac {4\pi}{\mu_0}\right)\dfrac {r}{2i}$$

Solution

The magnetic field due to a long current carrying conductor is given by:
$$|B|=\left( \dfrac {\mu_0}{4\pi}\right)\dfrac {2i}{r}$$