Finding the force on a charged particle in the earth's magnetic field EXAMPLE 24.8 The sun emits streams of charged particles (in what is called the solar wind) that move toward the earth at very high speeds. A proton is moving toward the equator of the earth at a speed of 500 km/s. At this point, the earth’s magnetic field is 5.0 × 10-5T SOLVE We use the steps of Tactics Box 24.2 to determine the force. O ỹ and B are perpendicular, so a = 90°. O The right-hand rule for forces tells us that the force will be directed parallel to the earth's surface. What are the direction and into the page in Figure 24.31. That is, the force is toward the the magnitude of the force on the proton? east. ® We compute the magnitude of the force using Equation 24.6: STRATEGIZE As we saw in Figure 24.7a, the field lines of the earth go from the earth's south pole to the earth's north pole; near the equator they are parallel to the earth's surface. Using this magnetic field direction and the direction of the proton's motion, we will use the steps of Tactics Box 24.2 to find the force on the F = |q|vB= (1.6 × 10-1ºC)(5.0 × 10° m/s)(5.0 × 10 5 T) = 4.0 × 10-18 N ASSESS This is a small force, but the proton has an extremely small mass of 1.67 × 10¯2' kg. Consequently, this force produces a very large acceleration: approximately 200 million times the acceleration due to gravity! proton. PREPARE FIGURE 24.31 shows a picture of the proton entering the earth's magnetic field. We need to convert the proton’s velocity to m/s: 1 × 10° m 500 km/s = 5.0 × 10² km/s X = 5.0 × 10° m/s 1 km FIGURE 24.31 A proton in the magnetic field of the earth. Your sketch will look more like this, with v and B in the plane of the paper. The force is into the page. The direction of the force on the proton is given by the right-hand rule. F into page

Glencoe Physics: Principles and Problems, Student Edition

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ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter25: Electromagnetic Induction

Section: Chapter Questions

Problem 71A

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In Example 24.8, we considered a proton in the solar wind moving toward the equator of the earth, where the earth’s magnetic field is 5.0 x 10^-5 T, at a speed of 500 km/s (5.0 x 10⁵ m/s). We now know that the proton will move in a circular orbit around the earth’s field lines. What are the radius and the period of the orbit?

Finding the force on a charged particle in the earth's magnetic field
EXAMPLE 24.8
The sun emits streams of charged particles (in what is called the
solar wind) that move toward the earth at very high speeds. A
proton is moving toward the equator of the earth at a speed of
500 km/s. At this point, the earth’s magnetic field is 5.0 × 10-5T
SOLVE We use the steps of Tactics Box 24.2 to determine the
force.
O ỹ and B are perpendicular, so a = 90°.
O The right-hand rule for forces tells us that the force will be
directed parallel to the earth's surface. What are the direction and
into the page in Figure 24.31. That is, the force is toward the
the magnitude of the force on the proton?
east.
® We compute the magnitude of the force using Equation 24.6:
STRATEGIZE As we saw in Figure 24.7a, the field lines of the
earth go from the earth's south pole to the earth's north pole; near
the equator they are parallel to the earth's surface. Using this
magnetic field direction and the direction of the proton's motion,
we will use the steps of Tactics Box 24.2 to find the force on the
F = |q|vB= (1.6 × 10-1ºC)(5.0 × 10° m/s)(5.0 × 10 5 T)
= 4.0 × 10-18 N
ASSESS This is a small force, but the proton has an extremely
small mass of 1.67 × 10¯2' kg. Consequently, this force produces
a very large acceleration: approximately 200 million times the
acceleration due to gravity!
proton.
PREPARE FIGURE 24.31 shows a picture of the proton entering the
earth's magnetic field. We need to convert the proton’s velocity
to m/s:
1 × 10° m
500 km/s = 5.0 × 10² km/s X
= 5.0 × 10° m/s
1 km
FIGURE 24.31 A proton in the magnetic field of the earth.
Your sketch will look more
like this, with v and B in the
plane of the paper. The force
is into the page.
The direction of the force
on the proton is given by
the right-hand rule.
F into page

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