A cosmic ray electron moves at 7.65 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where field strength is 1.10 ✕ 10−5 T. What is the radius (in m) of the circular path the electron follows?

Explanation:

Force of a magnetic field, Fm = q × v × B

Centipetal force, Fa = (M × v^2)/r

Therefore, Fm = Fa

q × v × B = (M × v^2)/r

r = (m × v)/(q × B)

Where,

r = radius

m = mass of electron

= 9.1 × 10^-31 kg

q = electron charge

= 1.602 × 10^-19 C

B = magnetic field

= 1.10 ✕ 10^−5 T

v = velocity

= 7.65×10^6 m/s

r = (9.1 × 10^-31) × (7.65 × 10^6) / (1.6 × 10^-19) × (1.10 × 10^-5)

= 3.95 m

c because the action was the roller coaster reaches bottom of hill (action) reaction would be blood tends to drain away from head

hope this !

ok

explanation: