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 4 of 8 pages

Question 31. The phase difference between the flux linkage and the induced e.m.f. in a rotating coil in a uniform magnetic field

Discuss Question

Answer: Option B. -> Π2
:
B

ϕ

=BAcos

θ

where

θ

is the angle between area and field

e = \frac{- d ϕ}{d t} = - B A s i n θ . \frac{d θ}{d t} = B A (\frac{d θ}{d t}) c o s (90 + θ)

Here phase difference =

90^{\circ} .

Question 32. A conducting ring is placed around the core of an electromagnet as shown in fig. When key K is pressed, the ring

Remain stationary
Is attracted towards the electromagnet
Jumps out of the core
None of the above

Discuss Question

Answer: Option C. -> Jumps out of the core
:
C
When key k is pressed, current through the electromagnet start increasing i.e. flux linked with ring increases which produces repulsion effect.

Question 33. Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B =

B_{0} e^{- t}

is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch is equal to

Shown In The Figure Is A Circular Loop Of Radius R And Resis...

B2oπ r2R
Bo10r3R
B2oπ2r4R5
B2oπ2r4R

Discuss Question

Answer: Option D. -> B2oπ2r4R
:
D

P = \frac{e^{2}}{R}; e = - \frac{d}{d t} (B A) = - A \frac{d}{d t} (B_{o} e^{- t}) = A B_{o} e^{- t}

\Rightarrow P = \frac{1}{R} (A B_{o} e^{- t})^{2} = \frac{A^{2} B_{o}^{2} e^{- 2 t}}{R}

At the time of starting t = 0 so

P = \frac{A^{2} B_{o}^{2}}{R}

\Rightarrow P = \frac{(π r^{2})^{2} B_{o}^{2}}{R} = \frac{B_{o}^{2} π^{2} r^{4}}{R}

Question 34. The resistance in the following circuit is increased at a particular instant. At this instant the value of resistance is 10W. The current in the circuit will be now

i = 0.5 A
i > 0.5 A
i < 0.5 A
i = 0

Discuss Question

Answer: Option B. -> i > 0.5 A
:
B
If resistance is constant

(10 Ω)

then steady current in the circuit

i = \frac{5}{10} = 0.5

A . But resistance is increasing it means current through the circuit start decreasing. Hence inductance comes in picture which induces a current in the circuit in the same direction of main current. So i

>

0.5 A.

Question 35. If a source is transmitting electromagnetic wave of frequency

8.2 \times 10^{6}

Hz, then wavelength of the electromagnetic waves transmitted from the source will be

36.6 m
40.5 m
42.3 m
50.9 m

Discuss Question

Answer: Option A. -> 36.6 m
:
A

λ = \frac{c}{v} = \frac{3 \times 10^{8}}{8.2 \times 10^{6}} = 36.5

Question 36. Which of the following represents an infrared wavelength

10−4
10−5
10−6
10−7

Discuss Question

Answer: Option A. -> 10−4
:
A
Wave length of infrared rays fall under the range

Question 37. If

⃗ E

and

⃗ B

are the electric and magnetic field vectors of E.M. waves then the direction of propagation of E.M. wave is along the direction of

⃗E
⃗B
⃗E× ⃗B
None of these

Discuss Question

Answer: Option C. -> ⃗E× ⃗B
:
C
EM waves travels with perpendicular to E and B. Which are also perpendicular to each other

⃗ v = ⃗ E \times ⃗ B

Question 38. Light is an electromagnetic wave. Its speed in vacuum is given by the expression

√μ0e0
√μ0e0
√e0μ0
1√μ0e0

Discuss Question

Answer: Option D. -> 1√μ0e0
:
D

μ_{0} = 4 π \times 10^{- 7}, ϵ_{0} = 8.85 \times 10^{- 12} \frac{N - m^{2}}{C^{2}}

c = \frac{1}{\sqrt{μ_{0} e_{0}}} = 3 \times 10^{8} \frac{m e t e r}{s e c} .

Question 39. In an apparatus, the electric field was found to oscillate with an amplitude of 18 V/m. The magnitude of the oscillating magnetic field will be