A charged particle is moving in a gravity free space in a uniform magnetic field. If an on instant of time velocity vector is V→=16i^-4j^m/s and acceleration vector is a→=i^-bj^m/s2 then magnitude of acceleration at that instant is
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a
23m/s2
b
4 m/s2
c
32 m/s2
d
5 m/s2
answer is D.
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Detailed Solution
Negative force F→=QV→×B→∴ acceleration a→=F→m=QmV→×B→Hence a→ is perpendicular to V→∴a→.V→=0⇒16-4×b=0⇒b = 4 ∴a→=32+42m/s2= 5 m/s2