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Question

The magnetic field of a plane electromagnetic wave is B=3×10-8sin200πy+cti^T wherec=3×108ms-1 is the speed of light . The corresponding electric field is


A

E=-9sin200πy+ctk^Vm

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B

E=9sin200πy+ctk^Vm

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C

E=-10-6sin200πy+ctk^Vm

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D

E=3×10-8sin200πy+ctk^Vm

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Solution

The correct option is A

E=-9sin200πy+ctk^Vm


Step 1: Given that:

The magnetic field of a plane electromagnetic wave is B=3×10-8sin200πy+cti^T

The speed of light c=3×108ms-1

Step 2: Formula used and draw the required diagram:

Know that an electromagnetic wave propagates in direction of the vector E×Band the phase gives the velocity of the wave.

The electric field vector of the EMwave is,

E=cB

Therefore

B=B0sin200πy+cti^B=3×10-8sin200πy+cti^E0=cB0E0=3×108×3×10-8

Therefore the value ofE0=9Vm

Know that the phase =y+ct is constant.

Then differentiate,

dydt=-c

Step 3: Solve the electric field E.

Note that the wave propagates in the negative y-direction.

E×Bis parallel to -j^.

E^×i^=λ-j^

Therefore,

E^=-k^

Hence, the electric field vector is,

E=E0sinky+ωt-k^E=-9sin200πy+ctk^Vm

Hence, the correct option is B.


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