Question

A proton of mass $1.67\times {10}^{-27}$ kg and charge $1.68\times {10}^{-19}$ C is projected with a speed of $2\times {10}^{6}m{s}^{-1}$ at an angle of ${60}^o$ to the $X-$axis. If a uniform magnetic field of $0.10$ T is applied along $Y-$axis, the path of proton is :

• 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 $4\pi \times {10}^{-7}s$

Answer 1

Answer: (C)

$v_x=v\cos \theta$
$v_y=v\sin \theta$
Since $v_y\ne 0$, this component will be responsible for motion along the direction $B($ i.e $y-$axis$)$.
The path will be helical and pitch will be constant.
Radius $r$ of the circle perpendicular to the magnetic field $B$

$r=\frac{mv\cos \theta }{qB}=\frac{1.6\times {10}^{-27}\times (2\times {10}^6\times \cos {60}^0)}{1.6\times {10}^{-19}\times 0.1}=0.1m$