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Physics Question on Electromagnetic waves

The magnetic field of a plane electromagnetic wave is given by B⃗=B0i^[cos(kz−ωt)]+B1j^cps(kz+ωt).\vec{B} = B_0 \hat{i} [cos (kz - \omega t)] + B_1 \hat{j} cps (kz + \omega t). where B0=3�10−5T_0 = 3 � 10^{-5} T and B1=2�10−6_1 = 2 � 10^{-6} T. The rms value of the force experienced by a stationary charge Q = 10 C at z = 0 is closest to :

A

0.9 N0.9 \,N

B

0.1 N0.1 \,N

C

3×10−2 N3 \times 10^-2\, N

D

0.6 N0.6\, N

Answer

0.6 N0.6\, N

Explanation

Solution

Maximum Electric field E = (B) (c) E0⃗=(3×10−5)c(−j^)\vec{E_0} = (3 \times 10^{-5})c (-\hat{j}) E1⃗=(3×10−6)c(−i^)\vec{E_1} = (3 \times 10^{-6})c (-\hat{i}) Maximum force F⃗net=qE⃗=qc(−3×10−5j^−2×10−6i^)\vec{F}_{net} = q\vec{E} = qc (-3 \times 10^{-5} \hat{j} - 2 \times 10^{-6} \hat{i}) F⃗0max = 10−4×3×108(3×10−5)2+(2×10−6)2\vec{F}_{0max} \, = \, 10^{-4} \times 3 \times 10^8 \sqrt{(3 \times 10^{-5})^2 + (2 \times 10^{-6})^2} =0.9N= 0.9 N Frms=F02 = 0.6N        (approx)F_{rms} = \frac{F_0}{\sqrt{2}} \, = \, 0.6 N \, \, \, \, \, \, \, \, (approx) Option (4)