Conceptual: 1, 3, 5, 6, 8, 16, 18, 19
Problems: 4, 6, 8, 11, 16, 20, 23, 27, 34, 41, 45, 56, 60, 65
Conceptual Questions
1. The magnetic field cannot be described as the magnetic force per unit charge because unlike the
electric force, the magnetic force depends upon the velocity of the charge.
3.
For a horizontal beam of electrons to be deflected to the right by a uniform magnetic field, the
field must point in the downward direction. With this orientation, the magnetic force would be to the left
for positively charged particles and to the right for those with a negative charge.
5.A constant magnetic field does zero work on a moving charge and therefore cannot change its kinetic
energy—the speed of a particle with constant kinetic energy must also be constant.
6. (a) The electron must be moving up or down along the direction of the magnetic field.
(b) The electron must be moving in the horizontal plane perpendicular to the magnetic field.
8.
The magnitude of the magnetic force is directly proportional to the charge of the moving particle.
Thus, a particle that is doubly charged experiences twice the magnetic force as one that is singly
charged. The radius of an object in centripetal orbit is equal to the ratio of the square of the orbital
velocity to the magnitude of the centripetal acceleration. The magnetic force does not change the
velocity of either charge, and thus, it is only the value of the acceleration that determines the radius. The
doubly-charged particle experiences a larger acceleration, and therefore, has a radius half the size of the
singly-charged particle.
16.(a) The picture on a TV or monitor is produced by high-speed electrons colliding with phosphors in
the screen. The electrons begin in the back of the picture tube and are accelerated and
steered by electromagnetic fields toward the screen. If a magnet is held near the screen, the
electrons are subject to an additional magnetic force that makes them change their course
and hit the screen in the wrong place. This causes the distortion in the picture.
(b) The metal mask behind the screen is made of a ferromagnetic material. If a magnet is held
near the screen the metal mask will be magnetized and some of the induced magnetization
may remain even after the magnet is removed. The picture will then remain distorted.
18. The magnetic force on a charge is always perpendicular to the charge’s velocity. Since work is
given by the applied force times the component of the displacement parallel to the force, the work
done by the magnetic force on a charged particle is always zero. Since the work done is zero, the
particle’s kinetic energy does not change. Therefore, the speed of a point charge acted on only by
a magnetic field is constant.
Problems
4.
Strategy A bar magnet is a magnetic dipole. Field lines emerge from a bar magnet at its north
pole and enter at its south pole. Magnetic field lines are closed loops. Use symmetry.
6.
Strategy A bar magnet is a magnetic dipole. Field lines emerge from a bar magnet at its north
pole and enter at its south pole. Magnetic field lines are closed loops. Use symmetry.
8. Strategy The magnetic force on a moving point charge is given by Eq. (19-5). The direction of the
force is determined by the right-hand rule and the sign of the charge.
Solution Find the magnetic force exerted on the electron.
v
e−
N
B
11. (a) Strategy Use conservation of energy.
Solution Find the speed of the electrons.
(b) Strategy Use Eq. (19-7).
Solution
(c) Strategy and Solution A magnetic field does no work on a moving point charge, so the
kinetic energy and, thus, the speed do not change. Therefore,
(d) Strategy Set the magnitudes of the electric and magnetic forces equal and solve for E.
Solution Find the electric field strength.
(e) Strategy and Solution Since the force due to the magnetic field is always perpendicular to
the velocity of the electrons, it does not increase the electrons’ speed but only changes their
direction.
16.Strategy Since the magnetic field points south and the magnetic force is upward, the (negatively
charged) electron’s component of velocity perpendicular to the magnetic field points east.
Solution Find the angle.
20. Strategy Solve Eq. (19-7) for the radius of the circular motion.
Solution Find the required radius of the vacuum chamber.
so
23.
Strategy Use conservation of energy and Newton’s 2nd law.
Solution Find the speed of the ion.
Solve for the magnitude of the magnetic field. First use F=ma, where F=qvB=evB, which leads to:
27. Strategy Since the electron experiences zero net force from the electric and magnetic forces, the
forces are equal in magnitude and opposite in direction. Use Newton’s 2nd law and show each force.
This eventually leads to a canned equation in the book (19-10). However, you should be able to come
up with this on your own for quizzes and exams. The canned equation is essentially useless unless you
can show the physics that creates it.
Solution Compute the electron’s speed.
34.(a) Strategy The potential energy of the proton due to the potential difference is converted into
kinetic energy. Use conservation of energy.
Solution Find the speed of the proton.
(b) Strategy The magnetic force on a moving point charge is given by Eq. (19-5). The direction of
the force is determined by the right-hand rule and the sign of the charge.
Solution The magnetic field is out of the page and the velocity of the proton is to the east, so
the magnetic force on the proton is to the south. The net force on the proton is zero, since it
travels in a straight line, so the force on the proton due to the electric field must be to the
north and its magnitude must be equal to that of the magnetic force. Compute the magnitude
of the electric field.
The electric field is in the direction of the force. Therefore, the electric field is
(c) Strategy and Solution Referring to part (b), we know that the magnetic force on the proton is
to the south, so in Region 3, it will follow
(d) Strategy The magnetic force causes the proton to accelerate radially. Use Newton’s second
law.
Solution Compute the radius of the circular path traveled by the proton in Region 3.
41. Strategy The net force in the y-direction must be zero. Use Newton’s second law and Eqs. (1912a) and (19-12b).
Solution Find the current.
so
The magnetic force must be upward to oppose gravity, so according to the
RHR and
the current is
45.(a) Strategy The torque on a current loop is
In this case (maximum torque)
Solution Find B.
(b) Strategy Use Eq. (19-12a) and symmetry.
Solution The field points from the north pole to the south. Due to symmetry,
the faces of the poles and into the page for the right half of the loop;
So,
is parallel to
is perpendicular to
is directed downward. By similar reasoning, the force on the right half of the
loop is equal and opposite. Thus, the torque is clockwise.
56. Strategy Use the principle of superposition and the field due to a long straight current-carrying
wire, Eq. (19-14).
Solution
(a) The fields are equal in magnitude
and opposite in direction at point P, so
(b) Both fields are now directed downward at P.
60. Strategy Use Eq. (19-17).
Solution Compute the magnetic field well inside the solenoid.
65. Strategy Use Ampère’s law. Currents out of the page are positive and into the page are negative.
Solution Sum the currents enclosed within each loop.
(a)
So, the net current is
(b)
So, the net current is