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# A charged particle moves with a constant velocity (^i+^j)m/s in a magnetic field →B=(2^i+3^k) and uniform electric field →E=(a^i+b^j+c^k)N/C , then (assuming all quantities in S.I. unit):a=−3b=3c=−2a2+b2+c2=22

A
a=3
B
a2+b2+c2=22
C
b=3
D
c=2
Solution
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#### Since the charge particle is moving with constant velocity, its acceleration will be 0.Fnet=0Fel+Fmag=0In equilibrium, electric force = - magnetic forceSo, q→E=−q(→v×→B) or a^i+b^j+c^k=−(3^i−3^j−2^k) So, a=−3,b=3,c=2and a2+b2+c2=9+9+4=22

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