Saved Bookmarks
| 1. |
A cosmic ray proton enters the Earth’s magnetic field in a direction perpendicular to the field. If the speed of the proton is 2 × 10 m/s and B = 1.6 × 10-6 T, find the force exerted on the proton by the magnetic field. [Charge on a proton, e = 1.6 × 10-19 C] |
|
Answer» Data : v = 2 × 107 m/s, B = 1.6 × 10-6 T, e = 1.6 × 10-19 C The magnetic force on the proton is Fm = evB (∵ \(\vec v\) ⊥ \(\vec B\)) = (1.6 × 10-19 C)(2 × 107 m/s)(1.6 × 10-6 T) = 5.12 × 10-18 N [Note : The Earth’s magnetic field traps the charged particles in doughnut-shaped regions outside the atmosphere. These regions are called Van Allen radiation belts. Near the poles, charged particles from these belts enter the atmosphere and produce the awesome shimmering curtains of light called the aurora borealis (northern ! lights) and aurora australis (southern lights).] |
|