Question

A beam of protons enters a uniform magnetic field of $0.3T$ with a velocity of $4\times {10}^{5}m/s$ in a direction making an angle of ${60}^{\circ }$ with the direction of magnetic field. What will be the pitch of the helix formed by moving particles? (Given charge of proton $e=1.6\times {10}^{-19}C$, mass of proton $m=1.67\times {10}^{-27}kg$)

• A. $4.37m$
• B. $0.437m$
• C. $0.0437m$
• D. $0.00437m$

Answer 1

Answer: (C)

$0.0437m$

The pitch of the helix is the distance travelled by it in one revolution. It will be as shown in the figure.
The component of velocity which will change due to magnetic field will be $v_{\perp }$. This is the component which will be responsible for the circular motion of the particles.
$T=\frac{2\pi m}{qB}$
Putting the values as given in the question, we get $T=\frac{2\pi \times 1.67\times {10}^{-27}}{1.6\times {10}^{-19}\times 0.3}\approx 21.86\times {10}^{-8}s$
$v_{||}=4\times {10}^5\cos {60}^{\circ }=2\times {10}^{5}m/s$
Pitch= $v_{||}T=2\times {10}^5\times 21.86\times {10}^{-8}$
$=43.72\times {10}^{-3}\approx 0.043m$