Electromagnetic Waves And Induction(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

ELECTROMAGNETIC WAVES AND INDUCTION MCQs

Electromagnetic Waves, Electromagnetic Induction, Waves On A String

Total Questions : 73 | Page 3 of 8 pages

Question 21. TV waves have a wavelength range of 1-10 meter. Their frequency range in MHz is

30-300
3-30
300-3000
3-3000

Discuss Question

Answer: Option A. -> 30-300
:
A

v = \frac{C}{λ} \Rightarrow v_{1} = \frac{3 \times 10^{8}}{1} = 3 \times 10^{8}

Hz=300 MHz
and

v_{2} = \frac{3 \times 10^{8}}{10} = 3 \times 10^{7}

Hz=30 MHz

Question 22. According to Maxwell’s hypothesis, a changing electric field gives rise to

An e.m.f.
Electric current
Magnetic field
Pressure radiant

Discuss Question

Answer: Option C. -> Magnetic field
:
C
According to the Maxwell’s EM theory, the EM waves propagation contains electric and magnetic field vibration in mutually perpendicular direction. Thus the changing of electric field give rise to magnetic field.

Question 23. A radio receiver antenna that is 2 m long is oriented along the direction of the electromagnetic wave and receives a signal of intensity

5 \times 10^{- 16} W / m^{2}

. The maximum instantaneous potential difference across the two ends of the antenna is

1.23 mV
1.23 mV
1.23 V
12.3 mV

Discuss Question

Answer: Option A. -> 1.23 mV
:
A

I = \frac{1}{2} ϵ_{0} C E_{0}^{2}

\Rightarrow E_{0} = \sqrt{\frac{2 I}{ϵ_{0} C}} = \sqrt{\frac{2 \times 5 \times 10^{- 16}}{8.85}} = 0.61 \times 10^{- 6} \frac{V}{m}

Alsop

E_{0} = \frac{V_{0}}{d} \Rightarrow V_{0} = E_{0} d = 0.61 \times 10^{- 6} \times 2 = 1.23 μ V

Question 24. In an electromagnetic wave, the amplitude of electric field is 1 V/m. the frequency of wave is

5 \times 10^{14}

Hz. The wave is propagating along z-axis. The average energy density of electric field, in

J o u l e m^{3}

, will be

1.1× 10−11
2.2× 10−12
3.3× 10−13
4.4× 10−14

Discuss Question

Answer: Option B. -> 2.2× 10−12
:
B
Average energy density of electric field is given by

u_{e} = \frac{1}{2} ϵ_{0} E^{2} = \frac{1}{2} ϵ_{0} {(\frac{E_{0}}{\sqrt{2}})}^{2} = \frac{1}{4} ϵ_{0} E_{0}^{2}

= \frac{1}{4} \times 8.85 \times 10^{- 12} (1)^{2} = 2.2 \times 10^{- 12} J / m^{3} .

Question 25. A TV tower has a height of 100 m. The average population density around the tower is 1000 per

k m^{2}

. The radius of the earth is

6.4 \times 10^{6}

m. The population covered by the tower is

2× 106
3× 106
4× 106
6× 106

Discuss Question

Answer: Option C. -> 4× 106
:
C
Population covered =

2 π h R \times

Population density

= 2 π \times 100 \times 6.4 \times 10^{6} \times \frac{1000}{(10^{3})^{2}} = 4 \times 10^{6}

Question 26. A circular loop of radius R carrying current I lies in x-y plane with its centre at origin. The total magnetic flux through x-y plane is

Directly proportional to I
Directly proportional to R
Directly proportional to
Zero

Discuss Question

Answer: Option D. -> Zero
:
D
Circular loop behaves as a magnetic dipole whose one surface will beN-pole and another will beS-pole. Therefore magnetic lines a force emerges fromNwill meet atS. Hence total magnetic flux throughx-yplane is zero.

Question 27. A short-circuited coil is placed in a time-varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the wire radius halved, the electrical power dissipated would be

Halved
The same
Doubled
Quadrupled

Discuss Question

Answer: Option B. -> The same
:
B
Power

P = \frac{e^{2}}{R};

hence

e = - (\frac{d ϕ}{d t})

where

ϕ

=NBA

∴ e = - N A (\frac{d b}{d t})

Also

R \propto \frac{1}{r^{2}}

Where R = resistance, r = radius, l =Length

∴ P \propto \frac{N^{2} r^{2}}{l} \Rightarrow \frac{P_{1}}{P_{2}} = 1

Question 28. In series with 20 ohm resistor a 5 henry inductor is placed. To the combination an e.m.f. of 5 volt is applied. What will be the rate of increase of current at t = 0.25 sec

e
e−2
e−1
None of these

Discuss Question

Answer: Option C. -> e−1
:
C

i = i_{o} (1 - e^{- \frac{R t}{L}}) \Rightarrow \frac{d i}{d t} = i_{o} (\frac{R}{L}) e^{\frac{R t}{L}} = \frac{E}{L} e^{\frac{R t}{L}}

On putting values

\frac{d i}{d t} = \frac{1}{e} = e^{- 1} .

Question 29. As shown in the figure, P and Q are two coaxial conducting loops separated by some distance. When the switch S is closed, a clockwise current

I_{p}

flows in P (as seen by E) and an induced current

I_{Q 1}

flows in Q. The switch remains closed for a long time. When S is opened, a current

I_{Q 2}

flows in Q. Then the directions of

I_{Q 1}

and

I_{Q 2}

(as seen by E) are

As Shown In The Figure, P And Q Are Two Coaxial Conducting L...

Respectively clockwise and anticlockwise
Both clockwise
Both anticlockwise
Respectively anticlockwise and clockwise

Discuss Question

Answer: Option D. -> Respectively anticlockwise and clockwise
:
D
When switch S is closed magnetic field lines passing through Q increases in the direction from right to left. So, according to Lenz’s law induced current in Q i.e.

I_{Q 1}

will flow in such a direction so that the magnetic field lines dueto

I_{Q 2}

passes from left to right through Q. This is possible when

I_{Q 1}

flows in anticlockwise direction as seen by E. Opposite is the case when switch S is opened i.e.

I_{Q 2}

will be clockwise as seen by E.

Question 30. A rectangular loop with a sliding connector of length l = 1.0 m is situated in a uniform magnetic field B = 2T perpendicular to the plane of loop. Resistance of connector is r = 2