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=−3
b=3
c=−2
a2+b2+c2=22
A
a=−3
B
b=3
C
a2+b2+c2=22
D
c=−2
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Solution
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Since the charge particle is moving with constant velocity, its acceleration will be 0.
Fnet=0
Fel+Fmag=0
In equilibrium, electric force = - magnetic force
So, 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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