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Question

Physics Question on Electromagnetic waves

An EM wave propagating in x-direction has a wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm-1. Choose the correct equations for electric and magnetic field if the EM wave is propagating in vacuum:

A

Ey=60sin[π4×103(x3×108t)]j^Vm1Ey=60sin⁡[\frac{π}{4}×10^3(x−3×10^8t)]\hat{j}Vm^{−1}
Bz=2sin[π4×103(x3×108t)]k^TBz=2sin⁡[\frac{π}{4}×10^3(x−3×10^8t)]\hat{k}T

B

Ey=60sin[π4×103(x3×108t)]j^Vm1Ey=60sin⁡[\frac{π}{4}×10^3(x−3×10^8t)]\hat{j}Vm^{−1}
Bz=2×107sin[π4×103(x3×108t)]k^TBz=2× 10^{-7} sin⁡[\frac{π}{4}×10^3(x−3×10^8t)]\hat{k}T

C

Ey=2×107sin[π4×103(x3×108t)]j^Vm1Ey=2× 10^{-7}sin⁡[\frac{π}{4}×10^3(x−3×10^8t)]\hat{j}Vm^{−1}
Bz=60sin[π4×103(x3×108t)]k^TBz=60 sin⁡[\frac{π}{4}×10^3(x−3×10^8t)]\hat{k}T

D

Ey=2×107sin[π4×104(x4×108t)]j^Vm1Ey=2× 10^{-7}sin⁡[\frac{π}{4}×10^4(x−4×10^8t)]\hat{j}Vm^{−1}
Bz=60sin[π4×104(x4×108t)]k^TBz=60 sin⁡[\frac{π}{4}×10^4(x−4×10^8t)]\hat{k}T

Answer

Ey=60sin[π4×103(x3×108t)]j^Vm1Ey=60sin⁡[\frac{π}{4}×10^3(x−3×10^8t)]\hat{j}Vm^{−1}
Bz=2×107sin[π4×103(x3×108t)]k^TBz=2× 10^{-7} sin⁡[\frac{π}{4}×10^3(x−3×10^8t)]\hat{k}T

Explanation

Solution

In first 3 options speed of light is 3 × 108 m/sec and in the fourth option it is 4 × 108 m/sec.
Using E = CB
We can check the option is B.