Question

A uniform magnetic field of 0.50 T is directed along the positive x axis. A proton moving with a speed of 60 km s enters this field. The helical path followed by the proton shown has a pitch of 5.0 mm. Determine the angle between the magnetic field and the velocity of the proton.

Answer 1

Explanation:

It is given that,

Magnetic field, B = 0.5 T

Speed of the proton, v = 60 km/s = 60000 m/s

The helical path followed by the proton shown has a pitch of 5.0 mm, p = 0.005 m

We need to find the  angle between the magnetic field and the velocity of the proton. The pitch of the helix is the product of parallel component of velocity and time period. Mathematically, it is given by :

$p=v_{||}\times T$

$p=v\ cos\theta\times \dfrac{2\pi m}{Bq}$

$cos\theta=\dfrac{pBq}{2\pi mv}$

$cos\theta=\dfrac{0.005\times 0.5\times 1.6\times 10^{-19}}{2\pi \times 1.67\times 10^{-27}\times 60000}$

$\theta=50.58^{\circ}$

So, the angle between the magnetic field and the velocity of the proton is 50.58 degrees. Hence, this is the required solution.