question_answer

A proton of mass \[1.67\times {{10}^{-27}}kg\] and charge \[1.6\times {{10}^{-19}}\,C\] is projected with a speed of \[2\times {{10}^{6}}\,m/s\] at an angle of \[60{}^\circ \] to the \[X-\]axis. If a uniform magnetic field of 0.104 Tesla is applied along \[Y-\]axis, the path of proton is  [IIT-JEE 1995]

A)            A circle of radius = 0.2 m and time period \[\pi \times {{10}^{-7}}s\]

B)            A circle of radius = 0.1 m and time period \[2\pi \times {{10}^{-7}}s\]

C)            A helix of radius = 0.1 m and time period \[2\pi \times {{10}^{-7}}s\]

D)            A helix of radius = 0.2 m and time period \[4\pi \times {{10}^{-7}}s\]

Correct Answer: C

Solution :

                   Path of the proton will be a helix of radius \[r=\frac{mv\sin \theta }{qB}\] (where q = Angle between \[\overrightarrow{B}\,\text{and}\,\overrightarrow{v\,}\]) \[\Rightarrow \ r=\frac{1.67\times {{10}^{-27}}\times 2\times {{10}^{6}}\times \sin {{30}^{o}}}{1.6\times {{10}^{-19}}\times 0.104}\] \[=0.1m\] Time period \[T=\frac{2\pi m}{qB}=\frac{2\pi \times 1.67\times {{10}^{-27}}}{1.6\times {{10}^{-19}}\times 0.104}\] \[=2\pi \times {{10}^{-7}}sec\]