A charged particle moves with a constant velocity (i^+j^)m/s in a magnetic field B=(2i^+3k^) and uniform electric field E=(ai^+bj^+ck^)N/C , then (assuming all quantities in S.I. unit):
This question has multiple correct options
A
a=−3
B
b=3
C
c=−2
D
a2+b2+c2=22
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Updated on : 2022-09-05
Solution
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Correct options are A) , B) and D)
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, qE=−q(v×B)or ai^+bj^+ck^=−(3i^−3j^−2k^)
So, a=−3,b=3,c=2and a2+b2+c2=9+9+4=22
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