Question

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 ${60}^{\circ }$ with the X-axis. If a uniform magnetic field of $0.104T$ is applied along Y-axis, the path of the proton is

• A. A circle of radius $=4.6\times {10}^{-14}m$ and time period $9\times {10}^{-13}s$ approximately
• B. A circle of radius $=3\times {10}^{-14}m$ and time period $5\times {10}^{-13}s$ approximately
• C. A helix of radius $=9.6\times {10}^{-14}m$ and time period $6\times {10}^{-7}s$ approximately
• D. A helix of radius $=2\times {10}^{-14}m$ and time period $4\times {10}^{-13}s$ approximately

Answer 1

The correct option is C A helix of radius $=9.6\times {10}^{-14}m$ and time period $6\times {10}^{-7}s$ approximately

We can separate the velocity of photon into two components and see their effects individually. Let the horizontal component of velocity be $v_x$ and vertical component be $v_y$.


Force on a charged particle moving with velocity $\overrightarrow{v}$ in a magnetic field $\overrightarrow{B}$ is given by
$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

Due to the horizontal component $v_x$,
$\overrightarrow{F}=q(\overrightarrow{v_x}\times \overrightarrow{B})$
$=q{v}_{x}B\sin {90}^{\circ }\hat{n}=q{v}_{x}B$ along positive z-axis (using the right hand rule)

Hence due to the horizontal component of velocity, the photon will perform circular motion in the x-z plane.


Due to the vertical component $v_y$,
$\overrightarrow{F}=q(\overrightarrow{v_y}\times \overrightarrow{B})$
$=q{v}_{y}B\sin 0^{\circ }\hat{n}=0$ (because $v_y$ and $B$ are parallel)

Hence there is no force due to velocity in the vertical direction. So, the photon will move perform helical motion (due to the combination of circular motion in x-z plane and rectilinear motion in y direction)


Radius of charged particle performing circular motion due to velocity $v_x$
$r=\frac{m{v}_x}{qB}=\frac{mv\cos {60}^{\circ }}{qB}$
$=\frac{1.6\times {10}^{-27}\times 2\times {10}^{-6}\times \frac{1}{2}}{0.104\times 1.6\times {10}^{-19}}$
$\approx 9.6\times {10}^{-14}m$
and time period
$R=\frac{2\pi r}{v_x}=\frac{2\pi r}{v\cos {60}^{\circ }}$
$=\frac{2\pi \times 9.6\times {10}^{-14}}{2\times {10}^{-6}\times \frac{1}{2}}$
$\approx 6\times {10}^{-7}s$