A charged particle is fired at an angle θ to a uniform magnetic field directed along the x–axis. During its motion along a helical path, if the pitch of the helical path is equal to the maximum distance of the particle from the x– axis, then
A
cosθ=1π
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B
sinθ=1π
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C
tanθ=1π
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D
tanθ=π
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Solution
The correct option is Dtanθ=π
The velocity perpendicular to magnetic field provides the circular motion, therefore m(vsinθ)2R=qvBsinθ ⇒R=mvsinθqB Pitch: P=2R=2mvsinθqB........(1) (This is the maximum distance of the particle from the x-axis) Time period: T=2πRvsinθ=2πvsinθ(mvsinθqB) ⇒T=2πmqB Again, pitch can be written as P=2πmqB×vcosθ......(2) From (1) and (2), 2πmqB×vcosθ=2mvsinθqB ⇒sinθcosθ=π ⇒tanθ=π