Doubly-ionized particles with an atomic mass of 59.4 u are moving due N at a speed of 7.78 x 105 m/s when they enter a uniform magnetic field of strength, B = 35.5 mT. The ions, which are moving perpendicular to the field, follow a circular path of radius, r, and travel north a distance, h = 2.02 m, before leaving the field at an angle, 8, as shown in the figure. (a) What is the radius of the ions' circular path? r = 6.793✔ m (b) At what angle do the ions leave the field? 0 = 73.44 X h 2

icon
Related questions
Question
Doubly-ionized particles with an atomic mass of 59.4 u are
moving due N at a speed of 7.78 x 105 m/s when they enter a
uniform magnetic field of strength, B = 35.5 mT.
The ions, which are moving perpendicular to the field, follow a
circular path of radius, r, and travel north a distance, h = 2.02 m,
before leaving the field at an angle, 8, as shown in the figure.
(a) What is the radius of the ions' circular path?
r = 6.793 m
(b) At what angle do the ions leave the field?
0 = 73.44 x
h
0
2
Transcribed Image Text:Doubly-ionized particles with an atomic mass of 59.4 u are moving due N at a speed of 7.78 x 105 m/s when they enter a uniform magnetic field of strength, B = 35.5 mT. The ions, which are moving perpendicular to the field, follow a circular path of radius, r, and travel north a distance, h = 2.02 m, before leaving the field at an angle, 8, as shown in the figure. (a) What is the radius of the ions' circular path? r = 6.793 m (b) At what angle do the ions leave the field? 0 = 73.44 x h 0 2
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

Blurred answer