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 8.03 x 106 m/s encounters the earth's magnetic field at an altitude where the field has a magnitude of 1.62 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 (a) an electron and (b) a proton.

Question

Physics

Coleman

28 July, 14:25

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 8.03 x 106 m/s encounters the earth's magnetic field at an altitude where the field has a magnitude of 1.62 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 (a) an electron and (b) a proton.

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Damarion White · Answer 1

Answer: a) r = 281m, b) r = 5.13*10^-7m

Explanation: From the question, we realised that the particle enters the earth magnetic field with it velocity perpendicular to the magnetic field thus making the particle have a circular motion.

The force exerted on a charge in a magnetic field perpendicular to the velocity of the particle is responsible for the centripetal force required to give the object it circular motion.

Magnetic force = centripetal force.

qvB = mv²/r

By dividing "v" on both sides, we have that

qB = mv/r

Above is the formulae that defines the circular motion of a particle in the earth's magnetic field

Where q = magnitude of electronic charge.

B = strength of magnetic field = 1.62*10^-7 T

v = speed of particle = 8.03*10^6 m/s

A) If the particle where to be an electron, q (magnitude of electron) = 1.609*10^-19c.

m = mass of electronic charge = 9.11*10^-31 kg.

By substituting these parameters into the formulae we have that

1.609*10^-19 * 1.62*10^-7 = 9.11*10^-31 * 8.03*10^6 / r

By cross multiplying, we have that

1.609*10^-19 * 1.62*10^-7 * r = 9.11*10^-31 * 8.03*10^6

r = 9.11*10^-31 * 8.03*10^6 / 1.609*10^-19 * 1.62*10^-7

r = 7.32*10^ (-24) / 2.61*10^-26

r = 2.81*10²

r = 281m

B)

If the particle is proton, q = magnitude of a proton charge = 1.609*10^-19c,

m = mass of proton = 1.673*10^-27 kg

By substituting these parameters into the formulae we have that

1.609*10^-19 * 1.62*10^-7 = 1.673*10^-27*8.03*10^6 / r

By cross multiplying, we have that

1.609*10^-19 * 1.62*10^-7 * r = 1.673*10^-27 * 8.03*10^6

r = 1.673*10^-27 * 8.03*10^6 / 1.609*10^-19 * 1.62*10^-7

r = 1.34*10^ (-32) / 2.61*10^-26

r = 0.513*10^-6 m

r = 5.13*10^-7m