Question

What velocity would a proton need to circle Earth 1,325 km above the magnetic equator, where Earth's magnetic field is directed horizontally north and has a magnitude of 4.00 ✕ 10⁻⁸ T? (Assume the raduis of the Earth is 6,380 km.)

____ magnitude in m/s

____ direction (either to the east, to the north, to the west, or to the south)

Answer 1

we will be using spherical polar coordinates.

Magnetic force on a charge is given by.

...(A)

Force required for the proton to circle the earth at height 1,325 Km is the centerpal force i.e

.......(B)

where r is the radius of circular orbit.Therefore r=(1325+6380) Km=7705*10^3 m.Therefore

At the magnetic equator B is horizantally north. i.e in the direction of θ thus we may write.

Thefore the magnetic force is given by,

Neglecting gravity, the centripetal force is equal to the magnetic force i.e

However

Comparing eqn (E) and (D)

from equations (C) and (F)

we have

And (- phi )is the direction towards the west.