hw12 - Physics 21 Fall, 2004 Solution to HW-12 27 P 29 An...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 21 Fall, 2004 Solution to HW-12 27 P29 An electron enters a uniform magnetic Feld 0.23 T ata45 to B . Determine the radius r and pitch p (distance between loops) of the electron’s helical path assuming its speed is 3 . 0 × 10 6 m/s. Since v makes an angle of 45 with B , the components of v parallel and perpendicular to B ( v ± and v ) both have the value v 0 / 2 (since sin θ =cos θ =1 / 2for θ =45 ), where v 0 is the given speed 3 . 0 × 10 6 m/s. We get the radius of the circular motion using v : r = m e v eB = (9 . 11 × 10 31 )(3 . 0 × 10 6 ) (1 . 6 × 10 19 ) 2(0 . 23) =52 . 5 µ m The time to make one loop of circular motion is t = 2 πr v = 2 πm e v eBv = 2 πm e eB . The pitch is the distance travelled parallel to B (at speed v ± ) in time ∆ t (one loop): p = v ± t = 2 πm e v ± eB =3 . 30 × 10 4 m = 330 µ m 27 P34 Show that the magnetic dipole moment µ of an electron orbiting the proton nucleus of a hydrogen atom is
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online