magnets and force

Question

What speed would a proton need to achieve 

in order to circle Earth 1050 km above the 

magnetic equator, where the Earth’s magnetic field is directed on a line between magnetic north and south and has an intensity of 

4.25 × 10−8 T? 

The mass of a proton is 1.673 × 10−27 kg. 

Answer in units of m/s.

Answer 1

The magnetic force F experienced by a charge q, moving with a speed v at an angle θ with the magnetic field B is given by:

F = qvBsinθ

In this case, θ=90° as v is perpendicular to B

Therefore F = qvB

For the situation described, i.e. a proton moving in a circle, the force qvB acts in the radial direction. Applying Newton's second law in the radial direction gives:

F_net = ma_c where a_c= centripetal acceleration = v²/r

i.e. F = mv²⁄r

since F = qvB, we have

qvB = mv²/r

i.e. qB = mv/r

solving for v yields: v = qBr/m

Substituting the values given: r = 1050 km = 1050000 m, q = 1.6 x 10 ^-19C, m =1.673 x 10^-27 kg, B = 4.25 x 10^-8T

v = 1.6 x 10 ^-19C* 4.25 x 10^-8T * 1050000m/1.673 x 10^-27 kg

v =4,267,782.43 C.T.m/kg

Units: (from F = qvB, 1 N = 1C.m/s.T and from F_net = ma, 1N = 1 kg.m/s² i.e. 1T =(1N/C).s/m = 1kg/C.s i.e. C.T.m/kg = m/s)

Therefore v = 4.27 x 10⁶ m/s (3 significant figures)