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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10-3 T in z direction

10-15 T in z−direction

6.25 ×10-3 T in z−direction

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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