The magnetic field of a plane electromagnetic wave is given by B=B_0 i[cos(kz-ω t)]+B_1 jcos(kz+ω t), where B_0=3× 10^–5 T and B_1=2× 10^–6 T. The…
JEE Main 2019 — Physics Electromagnetism
2019mcqhard
The magnetic field of a plane electromagnetic wave is given by B=B0i^[cos(kz−ωt)]+B1j^cos(kz+ωt), where B0=3×10–5T and B1=2×10–6T. The RMS value of the force experienced by a stationary charge Q=10–4C at z=0 is closest to
Official previous-year question
Held on 9 Apr 2019 · Verified 6 Jul 2026.
Options
A
0.1N
B
0.9N
C
3×10–2N
D
0.6N
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Solution
B=B0i^[cos(kz−ωt)]+B1j^cos(kz+ωt)
{B}_{0}=3\times {10}^{-5} T &{B}_{1}=2\times {10}^{-6}T
Electric filed associated with it is,
E=−B0cj^cos(kz−ωt)−B1ci^cos(kz+ωt)
Here, c is the speed of light in vacuum.
At z=0,
Frms=2(QB0c)2+(QB1c)2
Here c is speed of light in vacuum.
=2(10−4×3×10−5×3×108)2+(10−4×2×10−6×3×108)2
=20.81+0.0036=0.6N
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