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Q.

If a particle of charge 10−12  C moving along the x-direction with a velocity 105 m/s experiences a force of 10−10 newton in y-direction due to a magnetic field, then the minimum magnitude of magnetic field is

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a

10-3 T in  z direction

b

10-15 T  in  z−direction

c

6.25  ×10-3 T  in  z−direction

d

10-3 T in negative z direction

answer is D.

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Detailed Solution

F=qvB sin θ  ⇒  B=Fqv sin θBmin =Fqv  when θ=900 ∴  Bmin =Fqv  =  10−1010-12  ×  105 =  10−3  T   in negative z direction .
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