The solar wind is a thin, hot gas given off by the sun. Charged particles in this gas enter the magnetic field of the earth and can experience a magnetic force. Suppose a charged particle traveling with a speed of 9.80 x 10^6 m/s encounters the earth's magnetic field at an altitude where the field has a magnitude of 1.10 x 10^-7 T. Assuming that the particle's velocity is perpendicular to the magnetic field, find the radius of the circular path on which the particle would move if it were each of the following.

(a) an electron
_________ m

(b) a proton
____________m

please help me! this is due at midnight tonight!! thanks

1 answer

To solve this problem, we can use the following formula for the radius of the circular path of a charged particle in a magnetic field:

r = (m*v)/(q*B)

where r is the radius of the circular path, m is the mass of the particle, v is the speed of the particle, q is the charge of the particle, and B is the magnetic field strength.

(a) For an electron:
m = 9.11 x 10^-31 kg (mass of an electron)
q = 1.60 x 10^-19 C (charge of an electron)
v = 9.80 x 10^6 m/s (given)
B = 1.10 x 10^-7 T (given)

r_e = (9.11 x 10^-31 kg * 9.80 x 10^6 m/s) / (1.60 x 10^-19 C * 1.10 x 10^-7 T)
r_e ≈ 0.00051 m

(b) For a proton:
m = 1.67 x 10^-27 kg (mass of a proton)
q = 1.60 x 10^-19 C (charge of a proton)
v = 9.80 x 10^6 m/s (given)
B = 1.10 x 10^-7 T (given)

r_p = (1.67 x 10^-27 kg * 9.80 x 10^6 m/s) / (1.60 x 10^-19 C * 1.10 x 10^-7 T)
r_p ≈ 0.0877 m

So, the radius of the circular path on which the particle would move is approximately:

(a) 0.00051 m for an electron
(b) 0.0877 m for a proton