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Electromagnetic Waves question

2020 · 3 Sep · Shift 2 · Q48
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Electromagnetic Waves question

2020 · 3 Sep · Shift 2 · Q48

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
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)E=E0​j​cos(ωt−kx). The magnetic field B→\overrightarrow BB , at the moment t = 0 is :
  1. A
    B→=E0μ0∈0cos⁡(kx)j^\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat jB=μ0​∈0​​E0​​cos(kx)j​
  2. B
    B→=E0μ0∈0cos⁡(kx)k^\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat kB=μ0​∈0​​E0​​cos(kx)k
  3. C
    B→=E0μ0∈0cos⁡(kx)k^\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat kB=E0​μ0​∈0​​cos(kx)k
  4. D
    B→=E0μ0∈0cos⁡(kx)j^\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat jB=E0​μ0​∈0​​cos(kx)j​
View written solutionFree

Correct answer: C

  1. Given electric field

The wave is propagating along the +x+x+x direction, and

E⃗=E0j^cos⁡(ωt−kx)\vec E = E_0\hat j\cos(\omega t-kx)E=E0​j^​cos(ωt−kx)

So the electric field is along the yyy-axis.

  1. Direction of magnetic field

For an electromagnetic wave in vacuum:

  • E⃗⊥B⃗\vec E \perp \vec BE⊥B
  • both are perpendicular to the direction of propagation
  • E⃗×B⃗\vec E \times \vec BE×B gives the direction of propagation

Here,

  • propagation is along i^\hat ii^
  • E⃗\vec EE is along j^\hat jj^​

So we need j^×B⃗=i^\hat j \times \vec B = \hat ij^​×B=i^.

Using unit vectors,

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

Hence,

B⃗ is along k^.\vec B \text{ is along } \hat k.B is along k^.

So options A and D are immediately wrong.

  1. Magnitude relation between EEE and BBB

For an electromagnetic wave in vacuum,

E=cBE = cBE=cB

where

c=1μ0ϵ0c = \frac{1}{\sqrt{\mu_0\epsilon_0}}c=μ0​ϵ0​​1​

Thus,

B=Ec=Eμ0ϵ0B = \frac{E}{c} = E\sqrt{\mu_0\epsilon_0}B=cE​=Eμ0​ϵ0​​

So the amplitude of magnetic field is

B0=E0μ0ϵ0B_0 = E_0\sqrt{\mu_0\epsilon_0}B0​=E0​μ0​ϵ0​​

This eliminates option B and matches the coefficient in option C.

  1. Magnetic field at t=0t=0t=0

The full magnetic field corresponding to the given wave is

B⃗=B0k^cos⁡(ωt−kx)\vec B = B_0\hat k\cos(\omega t-kx)B=B0​k^cos(ωt−kx)

with

B0=E0μ0ϵ0B_0 = E_0\sqrt{\mu_0\epsilon_0}B0​=E0​μ0​ϵ0​​

At t=0t=0t=0,

B⃗=E0μ0ϵ0cos⁡(−kx)k^\vec B = E_0\sqrt{\mu_0\epsilon_0}\cos(-kx)\hat kB=E0​μ0​ϵ0​​cos(−kx)k^

Since cos⁡(−kx)=cos⁡(kx)\cos(-kx)=\cos(kx)cos(−kx)=cos(kx),

B⃗=E0μ0ϵ0cos⁡(kx)k^\vec B = E_0\sqrt{\mu_0\epsilon_0}\cos(kx)\hat kB=E0​μ0​ϵ0​​cos(kx)k^
  1. Compare with options

This matches:

B⃗=E0μ0ϵ0cos⁡(kx)k^\boxed{\vec B = E_0\sqrt{\mu_0\epsilon_0}\cos(kx)\hat k}B=E0​μ0​ϵ0​​cos(kx)k^​

which is Option C.

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