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.
A) 39°
B) 51°
C) 44°
D) 34°
E) 71°

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. 
A) 39° 
B) 51° 
C) 44° 
D) 34° 
E) 71°

Quizplus · Accepted Answer

We can use the formula for the radius of a charged particle moving in a magnetic field:

$r = \frac{mv}{qB}$

where $m$ is the mass of the particle, $v$ is its velocity, $q$ is its charge, and $B$ is the magnetic field strength.

For a proton in this case, we have:

$r = \frac{(1.67 	imes 10^{-27} kg)(60000 m/s)}{(1.60 	imes 10^{-19} C)(0.50 T)} = 3.12 	imes 10^{-3} m$

The pitch of the helix is given as 5.0 mm, which means that each complete turn of the helix corresponds to a vertical displacement of 5.0 mm. The vertical displacement is related to the radius of the helix by:

$	ext{pitch} = 2\pi r \sin	heta$

where $	heta$ is the angle between the velocity and the magnetic field.

Solving for $\sin	heta$, we get:

$\sin	heta = \frac{	ext{pitch}}{2\pi r} = \frac{(5.0 	imes 10^{-3} m)}{(2\pi)(3.12 	imes 10^{-3} m)} = 0.80$

Taking the inverse sine, we get:

$	heta = \sin^{-1}(0.80) = 51°$

Therefore, the angle between the magnetic field and the velocity of the proton is 51°, and the answer is (B).

A Uniform Magnetic Field of 0