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Question

# A point charge of 10C moves with a uniform velocity of 2^ax−4^azm/s in electromagnetic field having →E=^ax−3^ay+4^azV/mand^B=0.3^ax+0.1^ayWb/m2. The magnitude of force exerted on charge is

A
61.02 mN
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B
159.70 mN
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C
61.02 N
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D
159.70 N
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Solution

## The correct option is C 61.02 NNet force of charge is given as →F=→Fe+→Fm →F=q→E+q(→υ×→B) N →F=10[^ax−3^ay+4^az+(2^ax−4^az)×(0.3^ax+0.1^ay)] =10[^ax−3^ay+4^az+0.2^az−1.2^ay+0.4^ax] =10[1.4^ax−4.2^ay+4.2^az] →F=14^ax−42^ay+42^az ∣∣∣→F∣∣∣=√(14)2+(42)2+(42)2=61.02N

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