Question

At a certain location, the horizontal component of the earths magnetic field is 2.5 x 10^-5T, due north. A proton moves eastward with just the right speed for the magnetic force on it to balance its weight. Find the speed of the proton. e= 1.60x10^-19C m(p)=1.67x10^-27kg

Answer 1

here,

mass of proton , mp = 1.67 * 10^-27 kg

charge on proton , q = 1.6 * 10^-19 C

magnetic field , B = 2.5 * 10^-5 T

let the speed of electron be v

as the net force on proton is zero

magnetic force = gravitational force

q * v * B * sin(90) = m * g

1.6 * 10^-19 * v * 2.5 * 10^-5 = 1.67 * 10^-27 * 9.81

solving for v

v = 4.1 * 10^-3 m/s

the speed of proton is 4.1 * 10^-3 m/s