Question

A proton is circling the Earth above the magnetic equator, where Earth’s magnetic field is directed horizontally north and has a magnitude of 4.00 × 10–8 T. If the proton is moving at a speed of 2.7 × 107 m/s, how far above the surface of the Earth is the proton? Earth’s radius is 6.38 × 106 m, a proton’s mass is 1.67 × 10–27 kg, and its charge is 1.6 × 10–19 C.

a. 7.05E+6 m

b. 1.34E+7 m

c. 1.03E+8 m

d. 6.65E+5 m

Answer 1

here,

Magnetic equator is an img circle parallel to geograph equator.

let the height above the surface be h

for describing circular path under cross mag field (B in z diection and v, F and x-y plane parallel to equator

magnetic feild , B = 4 * 10^-8 T

speed , v = 2.7 * 10^7 m/s

radius of earth , r = 6.38 * 10^6 m

equating the forces

m * v^2 /(r + h) = q * v * B

1.67 * 10^-27 * (2.7 * 10^7)^2 /( 6.38 * 10^6 + h) = 1.6 * 10^-19 * 2.7 * 10^7 * 4 * 10^-8

solving for h

h = 6.65 * 10^5 m

the height above the surface is d) 6.65 * 10^5 m