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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

0.6N

B

0.1N

C

0.9N

D

`3 xx 10^(-2)N`

Text Solution

Verified by Experts

The correct Answer is:
a

`B_(0)= sqrt((B_(0))^(2)+ B_(1)^(2))= sqrt(30^(2)+ 2^(2)) xx 10^(-6)`
`~~ 30 xx 10^(-6)T`
`:. E_(0)= cB= 3 xx 10^(8) xx 30 xx 10^(-6)`
`= 9 xx 10^(3)V//m`
`E_(rms)= (E_(0))/(sqrt(2))= (9)/(sqrt(2)) xx 10^(3) V//m`
Force on the charge,
`F= E_(rms)Q= (9)/(sqrt(2)) xx 10^(3) xx 10^(-4) ~ 0.64N`
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