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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