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Electromagnetic Waves question

2020 · 6 Sep · Shift 2 · Q52
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  5. /2020 · 6 Sep · Shift 2 · Q52

Electromagnetic Waves question

2020 · 6 Sep · Shift 2 · Q52

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
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)B(x,t)=[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][1.2×10−7sin(0.5×103x+1.5×1011t)k] T The instantaneous electric field E→\overrightarrow EE corresponding to B→\overrightarrow BB is : (speed of light c = 3 × 108 ms–1)
  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]E(x,t)=[36sin(1×103x+1.5×1011t)i] Vm{V \over m}mV​
  2. 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}E(x,t)=[36sin(0.5×103x+1.5×1011t)k]mV​
  3. 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}E(x,t)=[36sin(1×103x+0.5×1011t)j​]mV​
  4. 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}E(x,t)=[−36sin(0.5×103x+1.5×1011t)j​]mV​
View written solutionFree

Correct answer: D

  1. Given magnetic field
B⃗(x,t)=[1.2×10−7sin⁡(0.5×103x+1.5×1011t) k^] T\vec B(x,t)=\left[1.2\times 10^{-7}\sin\left(0.5\times 10^3 x+1.5\times 10^{11}t\right)\,\hat k\right]\,\text{T}B(x,t)=[1.2×10−7sin(0.5×103x+1.5×1011t)k^]T

So:

  • Magnetic field amplitude: B0=1.2×10−7 TB_0=1.2\times 10^{-7}\,\text{T}B0​=1.2×10−7T
  • Direction of magnetic field: k^\hat kk^
  • Phase: 0.5×103x+1.5×1011t0.5\times 10^3 x+1.5\times 10^{11}t0.5×103x+1.5×1011t
  1. Find the direction of propagation

For a plane wave:

  • sin⁡(kx−ωt)\sin(kx-\omega t)sin(kx−ωt) means propagation in +x+x+x direction.
  • sin⁡(kx+ωt)\sin(kx+\omega t)sin(kx+ωt) means propagation in −x-x−x direction.

Here phase is:

kx+ωtkx+\omega tkx+ωt

so the wave propagates along negative x-direction, i.e. along −i^-\hat i−i^.

  1. Use the relation between E⃗\vec EE, B⃗\vec BB, and propagation direction

For an electromagnetic wave:

E⃗×B⃗\vec E \times \vec BE×B

gives the direction of propagation.

Since propagation is along −i^-\hat i−i^ and

B⃗∥k^,\vec B \parallel \hat k,B∥k^,

we need E⃗\vec EE such that

E⃗×k^=−i^\vec E \times \hat k = -\hat iE×k^=−i^

Now,

j^×k^=i^\hat j \times \hat k = \hat ij^​×k^=i^

therefore,

(−j^)×k^=−i^(-\hat j)\times \hat k=-\hat i(−j^​)×k^=−i^

Hence,

E⃗∥−j^\vec E \parallel -\hat jE∥−j^​
  1. Find the magnitude of electric field

For electromagnetic waves in vacuum:

E0=cB0E_0=cB_0E0​=cB0​

Given:

c=3×108 m/s,B0=1.2×10−7 Tc=3\times 10^8\,\text{m/s},\qquad B_0=1.2\times 10^{-7}\,\text{T}c=3×108m/s,B0​=1.2×10−7T

So,

E0=(3×108)(1.2×10−7)E_0=(3\times 10^8)(1.2\times 10^{-7})E0​=(3×108)(1.2×10−7) E0=3.6×101=36 V/mE_0=3.6\times 10^1=36\,\text{V/m}E0​=3.6×101=36V/m
  1. Phase of electric field

In an electromagnetic wave, E⃗\vec EE and B⃗\vec BB are in phase. Therefore E⃗\vec EE has the same argument:

0.5×103x+1.5×1011t0.5\times 10^3 x+1.5\times 10^{11}t0.5×103x+1.5×1011t

So,

E⃗(x,t)=[−36sin⁡(0.5×103x+1.5×1011t)j^]V/m\vec E(x,t)=\left[-36\sin\left(0.5\times 10^3 x+1.5\times 10^{11}t\right)\hat j\right] \text{V/m}E(x,t)=[−36sin(0.5×103x+1.5×1011t)j^​]V/m
  1. Compare with options
  • A: wrong phase and wrong direction
  • B: wrong direction (parallel to B⃗\vec BB, impossible)
  • C: wrong phase
  • D: exactly matches

Therefore, the correct option is D.

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