Electromagnetic Induction Test

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Electromagnetic Induction Test - 48

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Question 1
1 / -0

A straight line conductor of length $$0.4\ m$$ is moved with a speed of $$ 7 ms^{-1} $$ perpendicular to magnetic field of intensity of $$0.9\ Wbm^{-2} . $$ The induced emf across the conductor is :

A
$$25.2 \,V$$

B
$$5.04 \,V$$

C
$$2.52 \,V$$

D
$$1.26 \,V$$

Solution

The induced emf across the conductor $$E = Blv$$
$$ = 0.9 \times 0.4 \times 7 = 2.52 V $$
Question 2
1 / -0

Find the inductance L of a solenoid of length l whose windings are made of material of density D and resistivity $$ \rho . $$ The winding resistance is R :

A
$$ \dfrac {\mu_0}{4 \pi l} . \dfrac {R_m}{\rho D } $$

B
$$ \dfrac {\mu_0}{4 \pi R } . \dfrac {l_m}{\rho D } $$

C
$$ \dfrac {\mu_0}{4 \pi l } . \dfrac {R^2 m}{\rho D } $$

D
$$ \dfrac {\mu_0}{2 \pi R } . \dfrac {l_m}{\rho D } $$

Solution

For a solenoid $$ L = \mu_0 N^2 \dfrac {A}{l} 2 yx $$ is the length of the wire and a is the of cross-section, then
$$ R = \dfrac {px}{a} $$
and $$ m = axD $$
$$ R_m = \rho \dfrac {x}{a} = axD $$
$$ x = \sqrt{\dfrac{R_m}{\rho D }} $$
$$ x = 2 \pi r N , N = \dfrac {x}{2 \pi r} $$
$$ L = \mu_0 \left( \dfrac {x^2}{2 \pi r} \right)^2 \dfrac { \pi r^2}{l} $$
$$ = \dfrac {\mu_0}{4 \pi l} . \dfrac {R_m}{\rho D } $$
Question 3
1 / -0

When the self-inductance of the primary and secondary coil is doubled, then the mutual inductance of the two coils is :

A
Halved

B
Doubled

C
Quadrupled

D
Reduced to one-fourth

Solution

$$M=K\sqrt{L_1L_2}$$
If self-inductance is doubled, then mutual inductance is also doubled.
Question 4
1 / -0

Two coils $$A$$ and $$B$$ have mutual inductance $$2 \times {10}^{-2}$$ henry. If the current in the primary is $$i = 5 \sin{\left(10 \pi t\right)}$$ then the maximum value of e.m.f. induced in coil $$B$$ is

A
$$\pi\ volt$$

B
$$\dfrac{\pi}{2}\ volt$$

C
$$\dfrac{\pi}{3}\ volt$$

D
$$\dfrac{\pi}{4}\ volt$$

Solution

Current in the primary coil $$i = 5\sin (10 \pi t)$$
Rate of change of current $$\dfrac{di}{dt} = 50 \pi \cos(10\pi t)$$
Maximum rate of change of current $$\dfrac{di}{dt} \bigg|_{max}= 50 \pi$$
Mutual inductance between the coils $$L = 2\times 10^{-2}$$ H
Thus maximum emf induced $$\mathcal{E}_{max} = L\dfrac{di}{dt}\bigg|_{max}$$
$$\therefore$$ $$\mathcal{E}_{max} = 2\times 10^{-2 }\times 50\pi = \pi $$ volts
Question 5
1 / -0

Alternating current of peak value $$\left( \dfrac{2}{\pi} \right )$$ ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak e.m.f. induced in secondary coil is
(Frequency of a.c. = 50 Hz)

A
100 V

B
200 V

C
300 V

D
400 V

Solution

Given : $$I_o = \dfrac{2}{\pi}$$ ampere $$\nu = 50 $$ Hz $$L = 1 H$$
Thus $$w = 2\pi \nu = 2\pi (50) = 100 \pi$$
Alternating current flowing through the coil is given by $$I = I_o \sin wt$$
Differentiating it wr.t. time we get $$\dfrac{dI}{dt} = I_o w \cos wt$$
$$\therefore$$ $$\dfrac{dI}{dt}\bigg|_{max} = I_o w= \dfrac{2}{\pi} \times 100 \pi = 200$$ ampere per second
Peak e.m.f induced $$\mathcal{E} = L\dfrac{dI}{dt}$$
$$\implies$$ $$\mathcal{E} = 1\times 200 = 200$$ V
Question 6
1 / -0

If '$$N$$' is the number of turns in a circular coil then the value of self inductance varies as.

A
$$N^0$$

B
$$N$$

C
$$N^2$$

D
$$N^{-2}$$

Solution

Self-inductance of the coil is given by the relation
$$L = \dfrac{\mu_o N^2 A}{l}$$

here $$N$$ is the number of turns in the coil, $$A$$ and $$l$$ are the area and length of the coil, respectively.

$$\implies$$ $$L\propto N^2$$
Question 7
1 / -0

A conducting rod of mass $$m$$ and length $$l$$ is free to move without friction on two parallel long conducting rails, as shown below. There is a resistance $$R$$ across the rails. In the entire space around, there is a uniform magnetic field $$B$$ normal to the plane of the rod and rails. The rod is given an impulsive velocity $${ v }_{ 0 }$$. Finally, the initial energy $$\dfrac { 1 }{ 2 } m{ v }_{ 0 }^{ 2 }$$

A
Will be converted fully into heat energy in the resistor

B
Will enable rod to continue to move with velocity $${v}_{0}$$ since the rails are frictionless

C
Will be converted fully into magnetic energy due to induced current

D
Will be converted into the work done against the magnetic field

Solution

Magnetic force equals to $$F=ilB$$ ,will act opposite to the direction of $$v_o$$.The reduce in kinetic energy will be converted into heat energy by resistor.
Question 8
1 / -0

The figure shows a bar magnet coil. Consider four situations.
(I) Moving the magnet away from the coil.
(II) Moving the coil towards the magnet.
(III) Rotating the coil about the vertical diameter.
(IV) Rotating the coil about its axis.
An emf in the coil will be generated for the following situations.

A
(I) and (II) only

B
(I), (II) and (IV) only

C
(I), (II) and (III) only

D
(I), (II), (III) and (IV) only

Solution

To generate EMF we should have magnetic flux which changes with time. So when ever coil is moves away or close the flux will vary with time. Also rotating coil about vertical diameter will result in projected area and flux will change but when it is rotated about the axis then area and flux remain same .So the option 1,2 and 3 is correct.
Question 9
1 / -0

EMF developed by generator depends upon :

A
size of magnet

B
length of rotating wire

C
radius of wire

D
none of these

Solution

EMF developed by generator depends upon the length of rotating wire.An electric generator generates emf by changing the number of magnetic flux lines $$\phi$$ passing through a wire coil.When the coil is rotated between the poles of the magnet by cranking the handle,an AC voltage waveform is produced.
$$\varepsilon =-N\dfrac { d\phi }{ dt } $$
$$\varepsilon \alpha N$$ and $$N \alpha$$length of the coil
$$\varepsilon \alpha$$length of the coil.
Question 10
1 / -0

A wire of length $${l}$$ is moved with a constant velocity $$\vec{v}$$ in a magnetic field. A potential difference appears across the two ends.

A
If $$\vec{v}\|\vec{l}$$

B
If $$\vec{v}||\vec{B}$$

C
If $$\vec{l}||\vec{B}$$

D
None of these

Solution

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Electromagnetic Induction Test - 48

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Electromagnetic Induction Test - 48

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Question 1

$$25.2 \,V$$

$$5.04 \,V$$

$$2.52 \,V$$

$$1.26 \,V$$

Solution

Question 2

$$ \dfrac {\mu_0}{4 \pi l} . \dfrac {R_m}{\rho D } $$

$$ \dfrac {\mu_0}{4 \pi R } . \dfrac {l_m}{\rho D } $$

$$ \dfrac {\mu_0}{4 \pi l } . \dfrac {R^2 m}{\rho D } $$

$$ \dfrac {\mu_0}{2 \pi R } . \dfrac {l_m}{\rho D } $$

Solution

Question 3

Halved

Doubled

Quadrupled

Reduced to one-fourth

Solution

Question 4

$$\pi\ volt$$

$$\dfrac{\pi}{2}\ volt$$

$$\dfrac{\pi}{3}\ volt$$

$$\dfrac{\pi}{4}\ volt$$

Solution

Question 5

100 V

200 V

300 V

400 V

Solution

Question 6

$$N^0$$

$$N$$

$$N^2$$

$$N^{-2}$$

Solution

Question 7

Will be converted fully into heat energy in the resistor

Will enable rod to continue to move with velocity $${v}_{0}$$ since the rails are frictionless

Will be converted fully into magnetic energy due to induced current

Will be converted into the work done against the magnetic field

Solution

Question 8

(I) and (II) only

(I), (II) and (IV) only

(I), (II) and (III) only

(I), (II), (III) and (IV) only

Solution

Question 9

size of magnet

length of rotating wire

radius of wire

none of these

Solution

Question 10

If $$\vec{v}\|\vec{l}$$

If $$\vec{v}||\vec{B}$$

If $$\vec{l}||\vec{B}$$

None of these

Solution

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