A charged particle of mass m and charge q enters along AB at point A in a uniform magnetic field existing in the rectangular region of size a × b. The particle leaves the region exactly at corner point C. What is the speed v of the particle ?

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qBa2+b22mb

qBa2+b22ma

qBa2-b22ma

qBa2-b22mb

answer is A.

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Detailed Solution

Let speed of the particle be v (speed will remain constant).r=AO=CO=mvqB; Also ar = sin θ, r−br=cos θSolving the above equations, we get v=qBa2+b22mb

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