The solar wind is a thin, hot gas given off by the sun. Charged particles in thi

Question

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 8.60 106 m/s encounters the earth's magnetic field at an altitude where the field has a magnitude of 1.00 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.

Explanation / Answer

  Force on a charge (q) moving (at velocity v) perpendicular to a magnetic field (strength B) is
F = B . q . v

This force is at right angles to both the magnetic field and the direction of the velocity. Its is the same force that enables electric motors to work.

The force changes the direction of the velocity, but the new direction is still at right angles to the magnetic field, so the direction of the force is also shifted so that it is still at right angles to the velocity. This is easier to show with pictures but the end result of all the direction changing is that the charged particle performs circular motion.
If its going in a circle it needs a centripetal force. Thats the B.q.v

B.q.v must = m . v^2 ÷ r

One of the v's cancels and we can rearrange to get

r = (m . v) ÷ ( B . q)

You need to know the mass of the electron is 9.1 . 10^–31 kg

r = (9.1*10^–31*8.60*10^6 ) ÷ ( 1.0 *10^–7 *1.6 *10^–19 )

= 489.125 m ans A

the proton with mass 1.67 . 10^–27 kg

r = (1.67*10^–27 *8.60*10^6 ) ÷ ( 1.0 *10^–7 *1.6 *10^–19 )

=897625 m

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