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Electromagnetic Waves question

2017 · 9 Apr · Shift 1 · Q57
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  5. /2017 · 9 Apr · Shift 1 · Q57

Electromagnetic Waves question

2017 · 9 Apr · Shift 1 · Q57

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The electric field component of a monochromatic radiation is given by E→\overrightarrow EE= 2 E0 i^\widehat ii cos kz cos ω\omegaω t Its magnetic field B→\overrightarrow BB is then given by :
  1. A
    2E0cj^{{2{E_0}} \over c}\widehat jc2E0​​j​ sin kz cos ω\omegaω t
  2. B
    −2E0cj^-{{2{E_0}} \over c}\widehat j−c2E0​​j​ sin kz sin ω\omegaω t
  3. C
    2E0cj^{{2{E_0}} \over c}\widehat jc2E0​​j​ sin kz sin ω\omegaω t
  4. D
    2E0cj^{{2{E_0}} \over c}\widehat jc2E0​​j​ cos kz cos ω\omegaω t
View written solutionFree

Correct answer: B

  1. Given electric field

The electric field is

E⃗=2E0 i^ cos⁡kz cos⁡ωt\vec E = 2E_0 \, \hat i \, \cos kz \, \cos \omega tE=2E0​i^coskzcosωt

So only the xxx-component exists:

Ex=2E0cos⁡kzcos⁡ωtE_x = 2E_0 \cos kz \cos \omega tEx​=2E0​coskzcosωt
  1. Use Maxwell's equation

For electromagnetic waves,

∇×E⃗=−∂B⃗∂t\nabla \times \vec E = -\frac{\partial \vec B}{\partial t}∇×E=−∂t∂B​

Since E⃗=Exi^\vec E = E_x \hat iE=Ex​i^ and ExE_xEx​ depends only on zzz, the curl is

∇×E⃗=j^ ∂Ex∂z\nabla \times \vec E = \hat j \, \frac{\partial E_x}{\partial z}∇×E=j^​∂z∂Ex​​

with a minus sign from determinant evaluation:

∇×E⃗=j^(∂Ez∂x−∂Ex∂z)=−j^ ∂Ex∂z\nabla \times \vec E = \hat j \left(\frac{\partial E_z}{\partial x} - \frac{\partial E_x}{\partial z}\right) = -\hat j \, \frac{\partial E_x}{\partial z}∇×E=j^​(∂x∂Ez​​−∂z∂Ex​​)=−j^​∂z∂Ex​​

Now,

∂Ex∂z=2E0(−ksin⁡kz)cos⁡ωt\frac{\partial E_x}{\partial z} = 2E_0(-k\sin kz)\cos\omega t∂z∂Ex​​=2E0​(−ksinkz)cosωt

Therefore,

∇×E⃗=−j^ [−2E0ksin⁡kzcos⁡ωt]=2E0ksin⁡kzcos⁡ωt j^\nabla \times \vec E = -\hat j\,[ -2E_0 k \sin kz \cos\omega t] = 2E_0 k \sin kz \cos\omega t\, \hat j∇×E=−j^​[−2E0​ksinkzcosωt]=2E0​ksinkzcosωtj^​

Thus,

−∂B⃗∂t=2E0ksin⁡kzcos⁡ωt j^-\frac{\partial \vec B}{\partial t} = 2E_0 k \sin kz \cos\omega t\, \hat j−∂t∂B​=2E0​ksinkzcosωtj^​

so

∂B⃗∂t=−2E0ksin⁡kzcos⁡ωt j^\frac{\partial \vec B}{\partial t} = -2E_0 k \sin kz \cos\omega t\, \hat j∂t∂B​=−2E0​ksinkzcosωtj^​
  1. Integrate with respect to time

Let B⃗=Byj^\vec B = B_y \hat jB=By​j^​. Then

∂By∂t=−2E0ksin⁡kzcos⁡ωt\frac{\partial B_y}{\partial t} = -2E_0 k \sin kz \cos\omega t∂t∂By​​=−2E0​ksinkzcosωt

Integrating w.r.t. ttt,

By=−2E0ksin⁡kz∫cos⁡ωt dtB_y = -2E_0 k \sin kz \int \cos\omega t \, dtBy​=−2E0​ksinkz∫cosωtdt By=−2E0ksin⁡kz⋅sin⁡ωtωB_y = -2E_0 k \sin kz \cdot \frac{\sin\omega t}{\omega}By​=−2E0​ksinkz⋅ωsinωt​ By=−2E0kωsin⁡kzsin⁡ωtB_y = -\frac{2E_0 k}{\omega} \sin kz \sin\omega tBy​=−ω2E0​k​sinkzsinωt

For electromagnetic waves,

ωk=c⇒kω=1c\frac{\omega}{k} = c \quad \Rightarrow \quad \frac{k}{\omega} = \frac{1}{c}kω​=c⇒ωk​=c1​

Hence,

By=−2E0csin⁡kzsin⁡ωtB_y = -\frac{2E_0}{c} \sin kz \sin\omega tBy​=−c2E0​​sinkzsinωt

Therefore,

B⃗=−2E0c j^ sin⁡kz sin⁡ωt\boxed{\vec B = -\frac{2E_0}{c} \, \hat j \, \sin kz \, \sin\omega t}B=−c2E0​​j^​sinkzsinωt​
  1. Match with options

This corresponds to:

Option B\boxed{\text{Option B}}Option B​
  1. Check physically

Given

E⃗=2E0i^cos⁡kzcos⁡ωt\vec E = 2E_0\hat i \cos kz \cos\omega tE=2E0​i^coskzcosωt

This is a standing wave form, and the magnetic field should be 90∘90^\circ90∘ out of phase in time and proportional to sin⁡kzsin⁡ωt\sin kz\sin\omega tsinkzsinωt, with negative sign from Maxwell's equation. So Option B is consistent.

  1. Comparison with stored answer

Stored correct answer: C

But our derivation gives B, not C. Option C has the wrong sign.

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