Question

A cosmic ray electron moves at 7.50 × 106 m/s perpendicular to the Earth’s

magnetic field at an altitude where field strength is 1.00 × 10−5 T. What is the radius
of the circular path the electron follows?

Answer 1

4.27m

Step-by-step explanation:

force on moving charge= qvBsin$

centripetal force=mv^2/r

since the forces are equal

mv^2/r=qvBsin$

r= (mv^2)/(qvBsin$) = mv/qBsin$

r-radius

m-mass of an electron

q-charge on electron

v-velocity of the electron

B-field strength

$ - angle

r= ((9.1093837 ×10^-31)kg × 7.50 × 10^6 m/s)/(1.6×10^-19C×10^-5T×sin90) = 4.27m