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Electromagnetic Waves

JEE Physics · 127 questions · Page 4 of 13 · Click an option or "Show Solution" to reveal answer

Q31
The magnetic field of an electromagnetic wave is given by :- B→=1.6×10−6cos⁡(2×107z+6×1015t)(2i∧+j∧)Wbm2\mathop B\limits^ \to = 1.6 \times {10^{ - 6}}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){{Wb} \over {{m^2}}} The associated electric field will be :-
A E→=4.8×102cos⁡(2×107z−6×1015t)(−2i∧+j∧)Vm\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( -2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}
B E→=4.8×102cos⁡(2×107z−6×1015t)(2i∧+j∧)Vm\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( 2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}
C E→=4.8×102cos⁡(2×107z+6×1015t)(i∧−2j∧)Vm\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {\mathop i\limits^ \wedge - \mathop {2j}\limits^ \wedge } \right){V \over m}
D E→=4.8×102cos⁡(2×107z+6×1015t)(−i∧+2j∧)Vm\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( -{\mathop i\limits^ \wedge + \mathop {2j}\limits^ \wedge } \right){V \over m}
Correct Answer
Option C
Solution

If we use that direction of light propagation will be along

E→×B→\overrightarrow E \times \overrightarrow B

. Then (A) option is correct. Magnitude of E = CB E = 3 × 108 × 1.6 × 10–6 ×

5\sqrt 5

E = 4.8 ×

1025{10^{2\sqrt 5 }}
E→\overrightarrow E

and

B→\overrightarrow B

are perpendicular to each other

⇒E→.B→=0\Rightarrow \overrightarrow E .\overrightarrow B = 0

⇒\Rightarrow Either direction of

E→\overrightarrow E
i^−2j^\widehat i - 2\widehat j

or

−i^+2j^- \widehat i + 2\widehat j

from given option Also wave propagation direction is parallel to

E→×B→\overrightarrow E \times \overrightarrow B

which is

−k^- \widehat k

⇒\Rightarrow

E→\overrightarrow E

is along (

−i^+2j^- \widehat i + 2\widehat j

)

Q32
A plane electromagnetic wave travels in free space along the x-direction. The electric field component of the wave at a particular point of space and time is E = 6 V m–1 along y-direction. Its corresponding magnetic field component, B would be :
A 2 × 10–8 T along y-direction
B 6 × 10–8 T along z-direction
C 2 × 10–8 T along z-direction
D 6 × 10–8 T along x-direction
Correct Answer
Option C
Solution

The direction of propagation of an EM wave is direction of

E→×B→\overrightarrow E \times \overrightarrow B
i^=j^×B^\widehat i = \widehat j \times \widehat B
⇒B^=k^\Rightarrow \widehat B = \widehat k
C=EB⇒B=EC=63×108C = {E \over B} \Rightarrow B = {E \over C} = {6 \over {3 \times {{10}^8}}}

B = 2 × 10–8 T along z direction.

Q33
The mean intensity of radiation on the surface of the Sun is about 108 W/m2 . The rms value of the corresponding magnetic field is closet to :
A 102 T
B 10−-2 T
C 10−-4 T
D 1 T
Correct Answer
Option C
Solution

I =

ε0 C Erms2{\varepsilon _0}\,C\,E_{rms}^2

& Erms = cBrms I =

ε0{\varepsilon _0}

C3 B

rms2_{rms}^2

B

rms_{rms}

=

I∈0C3\sqrt {{{\rm I} \over {{ \in _0}{C^3}}}}

Brms ≈\approx 10−-4

Q34
For a plane electromagnetic wave, the magnetic field at a point x and time t is B→(x,t)\overrightarrow B \left( {x,t} \right) = [1.2×10−7sin⁡(0.5×103x+1.5×1011t)k^]\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right] T The instantaneous electric field E→\overrightarrow E corresponding to B→\overrightarrow B is : (speed of light c = 3 × 108 ms–1)
A E→(x,t)=[36sin⁡(1×103x+1.5×1011t)i^]\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {1 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat i} \right] Vm{V \over m}
B E→(x,t)=[36sin⁡(0.5×103x+1.5×1011t)k^]Vm\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]{V \over m}
C E→(x,t)=[36sin⁡(1×103x+0.5×1011t)j^]Vm\overrightarrow E \left( {x,t} \right) = \left[ {36\sin \left( {1 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}
D E→(x,t)=[−36sin⁡(0.5×103x+1.5×1011t)j^]Vm\overrightarrow E \left( {x,t} \right) = \left[ { - 36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}
Correct Answer
Option D
Solution

Given,

B→(x,t)\overrightarrow B \left( {x,t} \right)

=

[1.2×10−7sin⁡(0.5×103x+1.5×1011t)k^]\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]

T Wave is travelling along (–x) axis and

B→\overrightarrow B

is along +z axis.

We know, Magnitude of electric field E = BC = 1.2 ×\times 10-7 sin ( 0.5 10 + 1.5×\times1011t ) ×\times 3 ×\times 108 = 36 sin (0.5×\times103x+1.5×\times1011t) V/m Also,

s→=E→×B→μ0\overrightarrow s = {{\overrightarrow E \times \overrightarrow B } \over {{\mu _0}}}

⇒\Rightarrow

−i^=E→×k^μ0- \widehat i = {{\overrightarrow E \times \widehat k} \over {{\mu _0}}}

∴\therefore Direction of

E→=−j^{\overrightarrow E = - \widehat j}

∴\therefore Instantaneous electric field,

E→(x,t)=[−36sin⁡(0.5×103x+1.5×1011t)j^]Vm\overrightarrow E \left( {x,t} \right) = \left[ { - 36\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat j} \right]{V \over m}
Q35
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Electromagnetic waves carry energy but not momentum.Reason (R) : Mass of a photon is zero. In the light of the above statements, choose the most appropriate answer from the options given below :
A Both (A) and (R) are true and (R) is the correct explanation of (A)
B Both (A) and (R) are true but (R) is not the correct explanation of (A)
C (A) is false but (R) is true
D (A) is true but (R) is false
Correct Answer
Option C
Solution

Assertion is false because em waves have momentum.

Q36
The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is E→=E0j^cos⁡(ωt−kx)\overrightarrow E = {E_0}\widehat j\cos \left( {\omega t - kx} \right). The magnetic field B→\overrightarrow B , at the moment t = 0 is :
A B→=E0μ0∈0cos⁡(kx)j^\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat j
B B→=E0μ0∈0cos⁡(kx)k^\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat k
C B→=E0μ0∈0cos⁡(kx)k^\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat k
D B→=E0μ0∈0cos⁡(kx)j^\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat j
Correct Answer
Option C
Solution
E→=E0 cos⁡(ωt−kx)j^\overrightarrow E = {E_0}\,\cos (\omega t - kx)\widehat j

We know,

E=BCE = BC
B0=E0C=E01μ0∈0B_{0} = {E_{0} \over C} = {{{E_0}} \over {{1 \over {\sqrt {{\mu _0}{ \in _0}} }}}}

⇒\Rightarrow

B0=E0μ0∈0B_{0} = {E_0}\sqrt {{\mu _0}{ \in _0}}

You can see direction of

B→\overrightarrow B

is along z axis. ∴\therefore

B→=E0μ0∈0cos⁡(ωt−kx)k^\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos (\omega t - kx)\widehat k

at t = 0

B→=E0μ0∈0cos⁡(kx)k^\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos (kx)\widehat k
Q37
The electric fields of two plane electromagnetic plane waves in vacuum are given by E1→=E0j^cos⁡(ωt−kx)\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right) and E2→=E0k^cos⁡(ωt−ky)\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right) At t = 0, a particle of charge q is at origin with a velocity v→=0.8cj^\overrightarrow v = 0.8c\widehat j (c is the speed of light in vacuum). The instantaneous force experienced by the particle is :
A E0q(0.8i^−j^+0.4k^){E_0}q\left( {0.8\widehat i - \widehat j + 0.4\widehat k} \right)
B E0q(−0.8i^+j^+k^){E_0}q\left( { - 0.8\widehat i + \widehat j + \widehat k} \right)
C E0q(0.8i^+j^+0.2k^){E_0}q\left( {0.8\widehat i + \widehat j + 0.2\widehat k} \right)
D E0q(0.4i^−3j^+0.8k^){E_0}q\left( {0.4\widehat i - 3\widehat j + 0.8\widehat k} \right)
Correct Answer
Option C
Solution
E1→=E0j^cos⁡(ωt−kx)\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right)

Its corresponding magnetic field will be

B1→=E0ck^cos⁡(ωt−kx)\overrightarrow {{B_1}} = {{{E_0}} \over c}\widehat k\cos \left( {\omega t - kx} \right)
E2→=E0k^cos⁡(ωt−ky)\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right)

Also its corresponding magnetic field will be

B2→=E0ci^cos⁡(ωt−ky)\overrightarrow {{B_2}} = {{{E_0}} \over c}\widehat i\cos \left( {\omega t - ky} \right)

Net force on charge particle =

qE1→+qE2→+qv→×B1→q\overrightarrow {{E_1}} + q\overrightarrow {{E_2}} + q\overrightarrow v \times \overrightarrow {{B_1}}
+qv→×B2→+ q\overrightarrow v \times \overrightarrow {{B_2}}

=

qE0j^{q{E_0}\widehat j}

+

qE0k^{q{E_0}\widehat k}

+

q(0.8cj^)×(E0ci^)q\left( {0.8c\widehat j} \right) \times \left( {{{{E_0}} \over c}\widehat i} \right)

+

q(0.8cj^)×(E0ci^)q\left( {0.8c\widehat j} \right) \times \left( {{{{E_0}} \over c}\widehat i} \right)

=

E0q(0.8i^+j^+0.2k^){E_0}q\left( {0.8\widehat i + \widehat j + 0.2\widehat k} \right)
Q38
A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by B→=5×10−8j^T\overrightarrow B = 5 \times {10^{ - 8}}\widehat jT. The corresponding electric field E→\overrightarrow E is (speed of light c = 3 × 108 ms–1)
A 15 i^\widehat iV / m
B -15 i^\widehat iV / m
C 1.66 × 10–16 i^\widehat iV / m
D -1.66 × 10–16 i^\widehat iV / m
Correct Answer
Option A
Solution
E→=B→×V→\overrightarrow E = \overrightarrow B \times \overrightarrow V

=

(5×0−8j^)×(3×108k^)\left( {5 \times {0^{ - 8}}\widehat j} \right) \times \left( {3 \times {{10}^8}\widehat k} \right)

=

15 i^{15\,\widehat i}

V/m

Q39
The electric field of a plane electromagnetic wave is given by E→=E0i^+j^2cos⁡(kz+ωt)\overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right) At t = 0, a positively charged particle is at the point (x, y, z) = (0,0,πk)\left( {0,0,{\pi \over k}} \right). If its instantaneous velocity at (t = 0) is v0k^{v_0}\widehat k , the force acting on it due to the wave is :
A parallel to k^\widehat k
B parallel to i^+j^2{{\widehat i + \widehat j} \over {\sqrt 2 }}
C antiparallel to i^+j^2{{\widehat i + \widehat j} \over {\sqrt 2 }}
D zero
Correct Answer
Option C
Solution
E→=E0i^+j^2cos⁡(kz+ωt)\overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right)
E→\overrightarrow E

at t = 0 at z =

πk{\pi \over k}

is given by

E→=\overrightarrow E =
E0(i^+j^2)cos⁡(kπk+ω(0)){E_0}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)\cos \left( {k{\pi \over k} + \omega \left( 0 \right)} \right)

=

−E0(i^+j^2)- {E_0}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)

As

E→×B→=c→\overrightarrow E \times \overrightarrow B = \overrightarrow c

Force due to magnetic field is in direction

q(v→×B→)q\left( {\overrightarrow v \times \overrightarrow B } \right)

and given

v→{\overrightarrow v }

parallel to

k^\widehat k

.

F→=\overrightarrow F =
q(E→+v→×B→)q\left( {\overrightarrow E + \overrightarrow v \times \overrightarrow B } \right)

∴\therefore

F→\overrightarrow F

is in direction of

E→{\overrightarrow E }

which is antiparallel to

i^+j^2{{\widehat i + \widehat j} \over {\sqrt 2 }}
Q40
A plane electromagnetic wave of frequency 100 MHz is travelling in vacuum along the x-direction. At a particular point in space and time, B→=2.0×10−8k^T\overrightarrow B = 2.0 \times {10^{ - 8}}\widehat kT. (where, k^\widehat k is unit vector along z-direction) What is E→\overrightarrow E at this point?
A 0.6 j^\widehat j V/m
B 6.0 k^\widehat k V/m
C 6.0 j^\widehat j V/m
D 0.6 k^\widehat k V/m
Correct Answer
Option C
Solution

f = 100 MHz

B→=2×10−8T\overrightarrow B = 2 \times {10^{ - 8}}T
E→=B→×V→\overrightarrow E = \overrightarrow B \times \overrightarrow V
=(2×108k^)×(3×108i^)= (2 \times {10^8}\widehat k) \times (3 \times {10^8}\widehat i)
=6(k^×i^)=6j^= 6(\widehat k \times \widehat i) = 6\widehat j

V/m ⇒\Rightarrow

E→=6j^\overrightarrow E = 6\widehat j

V/m

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