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Q.

A beam of light travelling along x-axis is described by the electric field EY=(600Vm−1)sinω(t−xc),  then the maximum magnetic force on a charge q=2e moving along y-axis with a speed of 3×107m/s is (where e=1.6×10−19 C)

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a

1.92×10−16N

b

1.92×10−17N

c

1.6×10−16N

d

1.6×10−17N

answer is B.

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Detailed Solution

Amplitude of magnetic field is  B0=E0C=6003×108=2×10−6T( from the given equation,  E0=600V/m)Maximum magnetic force is  F=qvB0=2evB0 ⇒F=2×1.6×10−19×3×107×2×10−6 ⇒F=1.92×10−17N
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