Question

Hydrogen ion moving with a velocity of 10 for metre per second enters in magnetic field 10 power minus 4 tesla perpendicularly then its radius of circular path is

Answer 1

charge on hydrogen ion($H^+$), q = charge on a proton

= 1.6 × 10^-19C

mass of hydrogen ion = mass of proton

= 1.6726 × 10^-27 Kg

speed of hydrogen ion enters perpendicularly into magnetic field, v = 10 m/s

strength of magnetic field, B = 10^-4 T

radius of circular orbit, r = mv/qB

= (1.6726 × 10^-27 × 10)/(1.6 × 10^-19 × 10^-4)

= 1.6726 × 10^-3/1.6

= 1.04 × 10^-3 m

= 1.04 mm

hence, radius of circular path is 1.04 mm