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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a
qBa2+b22mb
b
qBa2+b22ma
c
qBa2-b22ma
d
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