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The magnetic field of a plane electromag...

The magnetic field of a plane electromagnetic wave is given by: `vec(B)=B_(0)hat(i)-[cos(kz- omegat)]+B_(1)hat(j)cos(kz+omegat)` where `B_(0)=3xx10^(-5)T` and `B_(1)=2xx10^(-6)T`. The rms value of the force experienced by a stationary charge `Q=10^(-4)C` at `z=0` is close to:

A

0.9 N

B

0.1N

C

0.6 N

D

` 3 xx 10^(-2) N`

Text Solution

Verified by Experts

The correct Answer is:
C
The electric field in the region is
`vec(E) = - c B_(0) cos ( omega t - kz) hat(j) - c B_(1) cos ( omega t + kz) hat(i)`
So for charge at rest at z = 0 the rms value of force is
`F_(rms) = Q sqrt(((cB_(0))/(sqrt(2)))^(2) + ((cB_(1))/(sqrt(2)))^(2))`
` = 10^(-4) xx ( 3 xx 10^(8))/( sqrt(2)) sqrt( (30 xx 10^(-6)) ^(2) + (2 + 10^(-6))^(2))`
`= 10^(-4) xx ( 3 xx10^(8))/( sqrt(2)) sqrt(904) xx 10^(-6) = 0.63` N
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