Question

A proton of speed v = 6 x 105 m/s enters a region of uniform magnetic field of B = 0.5 T at an angle of $ = 60° to the magnetic field. In the region of the magnetic field the proton describes a helical path with radius R and pitch p (distance between loops). Find R and p.

Answer 1

Solution

Radius of helix, R = 1.09cm

Pitch of the helix, rho = 3.954cm

The parallel component of velocity, $v\cos(\theta)$ will be unaffected, since there is no magnetic force acting parallel to the velocity. Hence this component of velocity gives the helical forward motion to the particle. It T is the time period taken to complete one complete rotation around the magnetic field, the pitch of the helix will be given by,

$\rho= v\cos(\theta)T$

Now the perpendicular component of velocity to the magnetic field will leads to magnetic force in the radial direction and hence a circular motion. The centripetal force required for circular motion will be given by the magnetic force due to the $v\sin(\theta)$ component. Thus,

$\\ \frac{mv^2\sin^2(\theta)}{r}=qv\sin(\theta)B \\ \\ \implies r=\frac{mv\sin(\theta)}{qB}$

Substituting the given values and using standard value of mass of proton, m=1.672*10^(-27)kg and charge, q=1.6*10^(-19)C

$\\ \implies r=\frac{1.672*10^{-27}*6*10^5\sin(60)}{1.6*10^{-19}*0.5}=0.0109m=1.09cm$Now using the relationship,$\\ v\sin(\theta)=r\omega=\frac{2\pi r}{T} \\ \\ \implies T=\frac{2\pi r}{v\sin(\theta)}=\frac{2*\pi*0.0109}{6*10^5*\sin(60)}=1.318*10^{-7}s$Thus pitch of helix,

$\rho=6*10^5\cos(60)*1.318*10^{-7}=0.03954m= 3.954cm$