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Question

Physics Question on Magnetic Field

A tightly wound long solenoid has 'nn' turns per unit length, a radius ' rr ' and carries a current II. A particle having charge ' qq ' and mass 'mm' is projected from a point on the axis in a direction perpendicular to the axis. The maximum speed of the particle for which the particle does not strike the solenoid is

A

μ0nIqrm\frac{\mu_{0} nIqr }{ m }

B

μ0nIqr2m\frac{\mu_{0} nIqr }{2 m }

C

μ0nIqr4m\frac{\mu_{0} nIqr }{4 m }

D

μ0nIqr8m\frac{\mu_{0} nIqr }{8 m }

Answer

μ0nIqr2m\frac{\mu_{0} nIqr }{2 m }

Explanation

Solution

FB=qvBsinθθ=90F _{ B } = qvB \sin \theta \,\, \theta = 90^{\circ} FB=qvBF _{ B } = qvB FC=mv2rF _{ C } =\frac{ mv ^{2}}{ r } FB=FCF _{ B } = F _{ C } qvB=mv2(r2)qvB =\frac{ mv ^{2}}{\left(\frac{ r }{2}\right)} v=qrB2mv =\frac{ qrB }{2 m } B=μ0nIB =\mu_{0} nI v=qrμ0nI2m v =\frac{ qr \mu_{0} nI }{2 m }