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Q.

A particle of specific charge α starts moving from the origin under the action of an electric field  E=E0i^ and magnetic field  B=B0k^. Its velocity at (x0,0,0) is (4i^+3j^). The value of x0 is 

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a

132αE0B0

b

16αB0E0

c

252αE0

d

5α2B0

answer is C.

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Detailed Solution

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Let q be the charge and m the mass of the particle. At (x0,0,0), speed of the particle is 5 units.
Applying work-energy theorem

qE0x0=12mv2=252mx0=25m2qE0=252αE0
 

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