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Question: If a particle of charge \(10 ^ { 5 } \mathrm {~m} / \mathrm { s }\) experiences a force of \(10 ^ { ...

If a particle of charge 105 m/s10 ^ { 5 } \mathrm {~m} / \mathrm { s } experiences a force of 10−1010 ^ { - 10 } newton in direction due to magnetic field, then the minimum magnetic field is

A

6.25×1036.25 \times 10 ^ { 3 } tesla in direction

B

in direction

C

6.25×10−36.25 \times 10 ^ { - 3 } tesla in direction

D

10−310 ^ { - 3 } tesla in direction

Answer

10−310 ^ { - 3 } tesla in direction

Explanation

Solution

F=qvBsin⁡θF = q v B \sin \theta ⇒ B=Fqvsin⁡θB = \frac { F } { q v \sin \theta }

Bmin⁡=FqvB _ { \min } = \frac { F } { q v } (when θ = 90o)

∴Bmin⁡=Fqv=10−1010−12×105=10−3\therefore B _ { \min } = \frac { F } { q v } = \frac { 10 ^ { - 10 } } { 10 ^ { - 12 } \times 10 ^ { 5 } } = 10 ^ { - 3 } Tesla in z^\hat { z } -direction.