Q. 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 $\times 10^{6}\ ms^{- 1}$ at an angle of 60${^\circ}$ to the X-axis. If a uniform magnetic field of 0.104 T is applied along Y-axis, the path of proton is
A) A circle of radius =0.2 m and time period
= $2\pi\ \times 10^{- 7}$ 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 = $2\pi\ \times 10^{- 7}$ s
Explanation
(c) Since, the proton is entering the magnetic field at some angle other than $90{^\circ}$, its path is helix. Component velocity of proto along $X$-axis,
$v_{x} = v\cos{60{^\circ} = 2 \times 10^{6} \times 1}$/2 = $10^{6}\ ms^{- 1}$
[image:4_image119.emf]
Due to component velocity $v_{x}$, the radius of the helix described is given by the relation
$r = \frac{mv_{x}}{qB} = \frac{1.67 \times 10^{- 27} \times 10^{6}}{1.6\ \times 10^{- 19} \times 0.104\ } =$0.1 m.
Now , $T = \frac{2\pi r}{v_{x}} = \frac{2\pi \times 0.1}{10^{6}} = 2\pi \times 10^{- 7}$s.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.