MAGNETIC FIELDS
A magnetic field is a region of space where a north magnetic monopole experiences a force. The direction of the field is by definition the direction of the force on the north end of a magnet. Since most texts contain diagrams of magnetic fields they will not be reproduced here.
CHARGES IN A MAGNETIC FIELD
Use your right hand to determine the direction of force on a moving positively charged particle in a magnetic field. With the fingers pointing from south to north (the same direction as the field), and the thumb pointing in the direction of the velocity of the particle, the palm points in the direction of the force on the particle. In the example below, the force is away from the observer.
For a negative particle, use your left hand.
The magnitude of the force on the particle, F, is
F = qvBsinø . Notice that F = 0 when any of the following are true:
q= 0 ; v = 0; ø = 0º; ø = 180º . Notice that F is greatest when the particle is moving at right angles to the field.
EXAMPLES
1. An alpha particle (two protons and two neutrons) traveling east at 2.0 x 10^{5} m/s enters a magnetic field of 0.20 T pointing straight up. What is the force acting on the alpha particle?
answer:
With the fingers of the right hand pointing straight up, and the thumb pointing east, the palm points south.
F = qvBsinø = (2 x 1.6 x 10^{19} C)(2.0 x 10^{5} m/s)(0.20 T)sin90º = 1.28 x 10^{14} N [S].
2. An electron traveling to the left, moves into a magnetic field directed toward the observer. Trace the path of the particle, assuming it eventually leaves the field.
answer:
The moment the electron enters the field, it experiences a force perpendicular to its velocity. The electron follows a circular path until it leaves the field.
FIELD AROUND A CURRENTCARRYING CONDUCTOR
When current flows through a conductor a magnetic field forms. The field lines form concentric circles around the conductor. Hold a straight conductor (wire), in your right hand with your thumb pointing in the direction of conventional current (positive flow). Your fingers circle the wire in the direction of the magnetic field. The compasses in the following diagram indicate the direction of the magnetic field near the conductor. Use your left hand for electron flow.
The following diagram shows a conductor carrying conventional current out of the page (toward the observer), and the direction of the field near the conductor. With the thumb of the right hand pointing toward the observer, the fingers of the right hand circle around the conductor in the direction of the magnetic field.
EXAMPLE
3. The following diagram is a view from above a table with a currentcarrying conductor stretched across the table top. Conventional current is flowing from left to right. A compass is laying flat on top of the conductor. What is the direction of the compass needle?
answer:
To the bottom of the page.
SOLENOIDS
Notice the fields formed by either side of a looped conductor reinforce each other inside the loop.
One might imagine carrying this a step further using many loops to produce a stronger magnetic field.
In practice, this is achieved by looping a wire many times and attaching it to one power source. This is called a "solenoid." An iron core is often used to help gather magnetic field lines into a more intense field. Hundreds or thousands of loops or "turns" are used, but only a few are diagrammed for clarity.
Notice that by wrapping the fingers of your right hand around the the solenoid in the direction of conventional current flow, the thumb points north. For electron flow use your left hand.
MAGNETIC FORCE ON A CURRENT
When a conductor carries current a current through a magnetic field, a magnetic force is produced on the conductor. (This is known as the motor principle). Use the right to determine the direction of magnetic force on a currentcarrying conductor in a magnetic field. The thumb points in the direction of conventional (positive) current. The fingers point in the direction of the field (from north to south). The palm points in the direction of the force on the conductor. In the diagram below, the force on the conductor is into the page (or away from the observer). For current defined as electron flow, use the left hand.
The magnitude of the force on the conductor, F, is
F = BILsinø where L is the length of the conductor in the field.
Notice that F = 0 when any of the following are true:
B= 0 ; I = 0; L = 0; ø = 0º; ø = 180º .
Notice that F is greatest when the conductor is at right angles to the field.
EXAMPLES
4. A horizontal conductor is carrying 5.0 A of current to the east. A magnetic field of 0.20 T pointing straight up cuts across 1.5 m of the conductor. What is the force acting on the conductor?
answer:
With the fingers of the right hand pointing straight up, and the thumb pointing east, the palm points south.
F = BILsinø = (0.20 T)(5.0 A)(1.5 m)sin90º = 1.5 N [S].
5. A 50.0 cm horizontal section of conductor with a mass of 8.00 g is in a 0.400 T magnetic field directed to the west. What are the magnitude and direction of current required to make this section of the conductor seem weightless?
answer:
The magnetic force must be opposite and equal to the weight of the section of the conductor.
With the fingers of the right hand pointing west, and the palm facing straight up, the thumb points north.
The weight of the conductor is mg = (0.00800 kg)(9.8 N/kg) = 0.0784 N .
The magnetic force on the conductor is
F = BILsinø , so
0.0784 N = (0.400 T)I(0.50 m)sin90º
I = 0.392 A [N]
ELECTROMAGNETIC INDUCTION
The previous section dealt with the magnetic force produced on a conductor carrying current in a field. This section describes electromagnetic induction: current caused to flow by forcing a conductor through a magnetic field.
Use your right hand to determine the direction of conventional current produced by moving a conductor through a magnetic field. With the fingers pointing in the direction of the magnetic field (from south to north), and the thumb pointing in the direction of the velocity of the conductor, the palm points in the direction of the current. In the example below, the current is into the paper (or away from the observer). For electron flow, use your left hand.
The magnitude of the electromotive force, or potential difference, is given by
V = BLvsinø , where L is the length of the conductor in the field. Notice that V = 0 when any of the following are true: B = 0 ; v = 0; l = 0; ø = 0º; ø = 180º . Notice that V is greatest when the conductor is moving at right angles to the field.
DOMAIN THEORY
The domain theory explains magnetic phenomenon by proposing the existence of domains. Domains are small regions within an object that are magnetic. These regions may be from one to hundreds of microns, which is small, but larger than atomic in size. When the polarities of the individual domains are randomized, their fields cancel one another and the object is not magnetic. When the polarities of all domains are parallel and aligned, their fields reinforce one another and the object is magnetic.
This theory explains several observations. When a ferrous material such as a nail is left in a magnetic field for an extended period of time the material becomes magnetic, at least temporarily. The explanation is that the domains are subject to atomic jostling and move randomly about fixed points. When exposed for a time to a magnetic field, the domains eventually line up with the field much as a compass needle lines up with the Earth's magnetic field. Once the domains are aligned, the object as a whole acts as a magnet. When removed from the field the jostling eventually randomizes the field again and the material becomes demagnetized.
Magnets become demagnetized when heated or when hammered or dropped repeatedly. This treatment randomizes the orientation of the domains which causes their individual fields to cancel.
PROBLEMS
(See below for answers)
1. An electron enters a 4.0 T field with a velocity of 5.0 x 10^{5} m/s perpendicular to the field. What is the radius of curvature of its path?
2. A conducting rod of 25 cm is pushed across a magnetic field along a Ushaped wire at a constant speed of 2.0 m/s. The field is directed away from the observer and is 4.00 T. A current of 8.00 A is induced in the circuit.
a. What is the potential difference induced in the circuit?
b. What is the resistance of the circuit?
c. What is the force used to push the rod?
d. In what direction is current flowing in the rod?
3. A wire with a linear density of 1 g/cm moves horizontally to the north on a horizontal surface with a coefficient of friction 0.2. What are the magnitude and the direction of the smallest magnetic field that enables the wire to continue in this fashion?
4. An electron is accelerated from rest through a potential difference of 18 kV, and then passes through a 0.34T magnetic field. Calculate the magnitude of the maximum magnetic force acting on the electron.
5. A 0.0017 T magnetic field and a 5.7 x 10^{3} N/C electric field both point in the same direction. A positive 2.0mC charge moves at a speed of 2.9 x 10^{6} m/s perpendicular to both fields. Determine the magnitude of the net force on the charge.
6. What is the magnitude of the magnetic force on an electron moving 5.0 x 10^{4} m/s perpendicular to a uniform magnetic field of .20T?
7. What speed would a proton need to orbit 1000 km above the Earth along the magnetic equator where the magnetic field intensity is 4.00x10^{8} T?
8.
a. Find the magnetic flux density 3.1 mm away from a long straight wire carrying a 1.2 A current.
b. What force per meter would act on a long straight wire 3.1mm away from and parallel to the wire in part (a) and carrying a 4.5A current?
c. What force would act on a 37 μC charged particle 3.1mm from the wire in part (a) and moving away from the wire at 6.5 m/s?
9. Find the magnetic flux density in the center of a 4.0 cm long aircore solenoid made with 4900 turns of wire and carrying a 2.5A current.
10. Express the attenuation distance for a plane electromagnetic wave in a good conductor in terms of the conductivity σ, permeability μ_{0}, and frequency ω.
ANSWERS
For solutions to all the problems on this page click here.
9.
For a long, thin coil, the magnetic flux density is
B = μ_{0}NI/L
where B is magnetic flux density (T); μ_{0} is the permeability of free space = 4π*10^{7} H/m; I is the current (A) and L is the length of the solenoid (m). For this problem,
B = (4π*10^{7})(4900)(2.5)/0.040) = 0.385 T
For solutions to all the problems on this page click here.
