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Q.

A particle with charge -5.60 nC is moving in a uniform magnetic field B→=−1.25Tk^ . The magnetic force on the particle is measured to be  F→=−(3.36×10−7N)i^+(7.42×10−7N)j^. Calculate one of the components of velocity of the particle from this information.

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a

−48 ms-1

b

4.8 ms-1

c

40.8 ms-1

d

−40.8 ms-1

answer is A.

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Detailed Solution

Let velocity of the particle is  v→=vxi^+vyj^+vzk^Then  F→=qv→×B→ ⇒ −3.36×10−7i^+7.42×10−7j^=−5.6×10−9vxi^+vyj^+vzk^×(−1.25)k^  ⇒ −3.36i^+7.42j^=7×10−2−vxj^+vyi^ ⇒ vx=−7.42×1027=−106 ms−1And  ⇒ vy=−3.36×1027=−48 ms−1.
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