<< Chapter < Page Chapter >> Page >

Strategy

  1. The direction of the magnetic field is shown by the RHR-1. Your fingers point in the direction of v , and your thumb needs to point in the direction of the force, to the left. Therefore, since the alpha-particles are positively charged, the magnetic field must point down.
  2. The period of the alpha-particle going around the circle is
    T = 2 π m q B .

    Because the particle is only going around a quarter of a circle, we can take 0.25 times the period to find the time it takes to go around this path.

Solution

  1. Let’s start by focusing on the alpha-particle entering the field near the bottom of the picture. First, point your thumb up the page. In order for your palm to open to the left where the centripetal force (and hence the magnetic force) points, your fingers need to change orientation until they point into the page. This is the direction of the applied magnetic field.
  2. The period of the charged particle going around a circle is calculated by using the given mass, charge, and magnetic field in the problem. This works out to be
    T = 2 π m q B = 2 π ( 6.64 × 10 −27 kg ) ( 3.2 × 10 −19 C ) ( 0.050 T ) = 2.6 × 10 −6 s.

    However, for the given problem, the alpha-particle goes around a quarter of the circle, so the time it takes would be
    t = 0.25 × 2.61 × 10 −6 s = 6.5 × 10 −7 s.

Significance

This time may be quick enough to get to the material we would like to bombard, depending on how short-lived the radioactive isotope is and continues to emit alpha-particles. If we could increase the magnetic field applied in the region, this would shorten the time even more. The path the particles need to take could be shortened, but this may not be economical given the experimental setup.

Check Your Understanding A uniform magnetic field of magnitude 1.5 T is directed horizontally from west to east. (a) What is the magnetic force on a proton at the instant when it is moving vertically downward in the field with a speed of 4 × 10 7 m/s? (b) Compare this force with the weight w of a proton.

a. 9.6 × 10 −12 N toward the south; b. w F m = 1.7 × 10 −15

Got questions? Get instant answers now!

Helical motion in a magnetic field

A proton enters a uniform magnetic field of 1.0 × 10 −4 T with a speed of 5 × 10 5 m/s . At what angle must the magnetic field be from the velocity so that the pitch of the resulting helical motion is equal to the radius of the helix?

Strategy

The pitch of the motion relates to the parallel velocity times the period of the circular motion, whereas the radius relates to the perpendicular velocity component. After setting the radius and the pitch equal to each other, solve for the angle between the magnetic field and velocity or θ .

Solution

The pitch is given by [link] , the period is given by [link] , and the radius of circular motion is given by [link] . Note that the velocity in the radius equation is related to only the perpendicular velocity, which is where the circular motion occurs. Therefore, we substitute the sine component of the overall velocity into the radius equation to equate the pitch and radius:

p = r v T = m v q B v cos θ 2 π m q B = m v sin θ q B 2 π = tan θ θ = 81.0 ° .

Significance

If this angle were 0 ° , only parallel velocity would occur and the helix would not form, because there would be no circular motion in the perpendicular plane. If this angle were 90 ° , only circular motion would occur and there would be no movement of the circles perpendicular to the motion. That is what creates the helical motion.

Got questions? Get instant answers now!
Practice Key Terms 2

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 2' conversation and receive update notifications?

Ask