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Q.

A proton of velocity 3i^+2j^ ms-1 enters a field of magnetic induction 2j^+3k^ tesla. The acceleration produced in the proton in ms-2 is (Specific charge of proton =0.96×108 C kg-1

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a

2.8×1082i^-3j^

b

2.88×1082i^-3j^+2k^

c

2.8×1082i^+3k^

d

2.88×108i^-3j^+2k^

answer is B.

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Detailed Solution

Here, v=3i^+2j^ ms-1 and B=2j^+3k^ T 

Specific charge of proton =em=0.96×108 C kg-1 

Force on a proton in a uniform magnetic field is 

F=ev×B=e3i^+2j^×2j^+3k^   =e6k^-9j^+6i^=e6i^-9j^+6k^N   

Acceleration of the proton, 

a=Fm=e6i^-9j^+6k^m=0.96×1086i^-9j^+6k^ ms-2   =2.88×1082i^-3j^+2k^ ms-2   

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