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Question

A plane electromagnetic wave of frequency 25MHz travels in free space along the x-direction. At a particular point in space and time, E=6.3j^Vm . What is B at this point?


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Solution

Step 1. Given data

Wave frequency =25MHz

Intensity of electric field E=6.3j^Vm

We have to find the magnitude of B.

Step 2. Formula to be used.

The magnitude of the electric field is simply defined as the force per charge on the test charge.

The magnitude of the electric field vector and magnetic field vector of the electromagnetic wave is related as,

EB=C

Here, C is the speed of light in vacuum.

Step 3. Find the value of B.

From the given, the value of E is 6.3 and C is 3×108m/s.

B=Ec

=6.33×108

=2.1×10-8

Now, as we know, for electromagnetic waves, E and B are perpendicular to each other as well as to the direction of propagation of wave E×B.

Where the direction of E is positive y- direction, means j^ and direction of propagation of the wave is in positive x- direction i. e. i^.

If B has to be perpendicular to E and direction of propagation, then it must in positive z- direction i. e. k^.

Therefore, the magnetic field is,

B=2.1×10-8k^T

Hence, the magnitude of B is 2.1×10-8k^T.


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