   Chapter 19, Problem 17P

Chapter
Section
Textbook Problem

Jupiter’s magnetic field occupies a volume of space larger than the Sun and contains ionized particles ejected from sources including volcanoes on Io, one of Jupiter’s moons. A sulfur ion (S+) in Jupiter’s magnetic field has mass 5.32 × 10−26 kg and kinetic energy 75.0 eV. (a) Find the maximum magnetic force on the ion from Jupiter’s magnetic field of magnitude 4.28 × 10−4 T. (b) Find the radius of the sulfur ion’s circular path, assuming its velocity is perpendicular to Jupiter’s magnetic field.

a)

To determine
The maximum magnetic force on the ion from Jupiter’s magnetic field.

Explanation

Given info: The mass of the sulfur ion in jupiter has mass 5.32×1026kg . The kinetic energy of the ions is 75.0eV . The magnitude of the magnetic field of Jupiter is 4.28×104T .

Explanation:

The kinetic energy is given by,

KE=12mv2

• m is the mass of the ion
• v is the velocity of the ion

Re-arranging for the velocity,

v=2(KE)m       (1)

The magnetic force on a charged particle is given by,

F=qvBsinθ

• q is the charge on the particle
• v is the velocity of the particle
• B is the magnetic field
• θ is the angle between the velocity and the magnetic field

Since the velocity of the proton is perpendicular to the earths magnetic field, the magnetic force will be maximum having a value,

F=qvB       (2)

From (1) and (2),

F=q(2(KE)m)B

Substitute 1.60×1019C for q , 5

b)

To determine
The radius of the sulfur ions circular path.

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