The magnetic field of a plane EM wave is given by B=Boi^[cos(kx−wt)]+B1j^cos(kz+wt) where B0=3×10−5T and B1=2×10−6T. The rms value of the force experienced by stationary charge Q=10−4C at z = 0 is closest to

On stationary charge no magnetic force will act.Maximum electric field E = (B) CE0→=(3×10−5)c(−j^) ; E1→=(2×10−6)c(−i^) Maximum force Fmax=qE→=qE02+E12Fmax=0.9 Frms=Fmax2=0.6

The magnetic field of a plane EM wave is given by B=Boi^[cos(kx−wt)]+B1j^cos(kz+wt) where B0=3×10−5T and B1=2×10−6T. The rms value of the force experienced by stationary charge Q=10−4C at z = 0 is closest to

On stationary charge no magnetic force will act.Maximum electric field E = (B) CE0→=(3×10−5)c(−j^) ; E1→=(2×10−6)c(−i^) Maximum force Fmax=qE→=qE02+E12Fmax=0.9 Frms=Fmax2=0.6

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The magnetic field of a plane EM wave is given by $B = B_{o} \hat{i} [\cos (k x - w t)] + B_{1} \hat{j} \cos (k z + w t)$
where $B_{0} = 3 \times 10^{- 5} T a n d B_{1} = 2 \times 10^{- 6} T .$ The rms value of the force experienced by stationary charge $Q = 10^{- 4} C$ at z = 0 is closest to

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0.6 N

0.9N

0.1N

$3 \times 10^{- 2} N$

answer is D.

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Detailed Solution

On stationary charge no magnetic force will act.

Maximum electric field E = (B) C
$\vec{E_{0}} = (3 \times 10^{- 5}) c (- \hat{j}); \vec{E_{1}} = (2 \times 10^{- 6}) c (- \hat{i})$
Maximum force $F_{\max} = q |\vec{E}| = q \sqrt{E_{0}^{2} + E_{1}^{2}}$
$F_{\max} = 0.9 F_{r m s} = \frac{F_{\max}}{\sqrt{2}} = 0.6$