At a certain location the horizontal component of the earths

At a certain location, the horizontal component of the earth\'s magnetic field is 2.9 × 10-5 T, due north. A proton moves eastward with just the right speed, so the magnetic force on it balances its weight. Find the speed of the proton.

Solution

The magnetic force on the proton is:
Fm = qv x B = q v x^ x By^ = (qvB) x^ x y^ = (qvB) z^

q = elementary unit of charge = 1.6e-19 (C)
B = 2.7e-5 (T)

The gravitational force on the proton is:
Fg = -mg z^

m = proton mass = 1.67e-27 (kg)
g = local acceleration of gravity = 9.8 (m/s^2)

Therefore, if the forces balance:
qvB z^ = Fm = - Fg = mg z^
or
qvB = mg
or
v = mg/(qB) = (1.67e-27)(9.8)/((1.6e-19)(2.9e-5))
= 3.53e-3 (m/s)

So v = 3.53*10^(-3) (m/s)