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Physics Question on Electromagnetic waves

A plane EM wave is propagating along x direction. It has a wavelength of 4 mm. If electric field is in y direction with the maximum magnitude of 60 V m-1, the equation for magnetic field is:

A

Bz=60sin⁡[π2(x−3×108t)]k^ TB_z = 60 \sin \left[ \frac{\pi}{2} \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \text{T}

B

Bz=2×10−7sin⁡[π2×103(x−3×108t)]k^ TB_z = 2 \times 10^{-7} \sin \left[ \frac{\pi}{2} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \text{T}

C

Bx=60sin⁡[π2(x−3×108t)]i^ TB_x = 60 \sin \left[ \frac{\pi}{2} \left( x - 3 \times 10^8 t \right) \right] \hat{i} \, \text{T}

D

Bz=2×10−7sin⁡[π2(x−3×108t)]k^ TB_z = 2 \times 10^{-7} \sin \left[ \frac{\pi}{2} \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \text{T}

Answer

Bz=2×10−7sin⁡[π2×103(x−3×108t)]k^ TB_z = 2 \times 10^{-7} \sin \left[ \frac{\pi}{2} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \text{T}

Explanation

Solution

Step 1: Relation between electric and magnetic fields

The relationship between the electric field EE and magnetic field BB is:

E=cB,E = cB,

where c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}.

Substitute E=60 Vm−1E = 60 \, \text{Vm}^{-1}:

60=3×108⋅B  ⟹  B=603×108=2×10−7 T.60 = 3 \times 10^8 \cdot B \implies B = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \, \text{T}.

Step 2: Calculate the frequency

The wavelength is given as:

λ=4 mm=4×10−3 m.\lambda = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m}.

The wave velocity cc is related to the frequency ff as:

c=fλ  ⟹  f=cλ=3×1084×10−3=34×1011 Hz.c = f\lambda \implies f = \frac{c}{\lambda} = \frac{3 \times 10^8}{4 \times 10^{-3}} = \frac{3}{4} \times 10^{11} \, \text{Hz}.

Step 3: Angular frequency

The angular frequency ω\omega is given by:

ω=2πf=2π⋅34×1011=3π2×1011.\omega = 2\pi f = 2\pi \cdot \frac{3}{4} \times 10^{11} = \frac{3\pi}{2} \times 10^{11}.

Thus:

ω=π2×103.\omega = \frac{\pi}{2} \times 10^3.

Step 4: Determine the direction of the fields

  • The electric field is in the yy-direction (ȷ^\hat{\jmath}).
  • The wave propagates in the xx-direction (ı^\hat{\imath}).
  • The magnetic field must be perpendicular to both ı^\hat{\imath} and ȷ^\hat{\jmath}, i.e., in the zz-direction (k^\hat{k}).

Step 5: Equation of the magnetic field

The magnetic field BzB_z is:

Bz=2×10−7sin⁡[π2×103(x−3×108t)]k^.B_z = 2 \times 10^{-7} \sin\left[\frac{\pi}{2} \times 10^3 \left(x - 3 \times 10^8 t\right)\right] \hat{k}.

Final Answer: Bz=2×10−7sin⁡[π2×103(x−3×108t)] k^ kT.B_z = 2 \times 10^{-7} \sin\left[\frac{\pi}{2} \times 10^3 \left(x - 3 \times 10^8 t\right)\right] \, \hat{k} \, \text{kT}.