We will skip the subsection on Ampère's Law.
A wire carrying a current produces a magnet field. A long straight wire carrying a current produces a magnetic field that is tangent to a circle centered on the wire and perpendicular to it. The direction of the magnetic field is given by a second right hand rule:
If the wire is grasped in the right hand with the thumb in the direction of the current, the fingers will curl in the direction of B.
The two wires in Figure P19.39 carry currents of 3.00A and 5.00A in the direction indicated. Find the direction and magnitude of the magnetic field at a pont midway between the wires.
Using the right hand rule #2, we find that, halfway between the wires, the magnetic field produced by the 3.00A wire points straight up and the magnetic field produced by the 5.00A wire points straight down.
The midway point is at r=10.0cm from each wire.
The resulting magnetic field is the vector sum of these two contributions:
Btot = B3A - B5A
where define up to be positive.
B3A = m0I/2pr = (4p×10-7Tm/A)(3.00A)/2p(0.100m) = 6.00×10-6T and
B5A = m0I/2pr = (4p×10-7Tm/A)(5.00A)/2p(0.100m) = 10.00×10-6T.
The total field is
Btot = (6.00 - 10.00)×10-6 T = -4.00×10-7T.
The negative sign tells us that the field points downward at that point.
Since a wire carrying a current creates a magnetic field, and a wire carrying a current feels a force when in a magnetic field, two wires carrying currents will exert a force between them. Let's consider two parallel conductors carrying currents I1 and I2, separated by a distance d, and consider the force exerted on wire 1 (draw sketch).
The force on wire 1 will be F1 = B2I1l sinq, where B2 is the magnetic field created by wire 2 at the location of wire 1, and q is the angle between wire 1 and the direction of B2.
Wire 2 creates a magnetic field that is tangent circles centered on wire 2. The magnetic field at wire 1 is tangent to a circle of radius d. Therefore, the lines of magnetic field are perpendicular to wire 1, q = 90°, and sinq = 1. This is true where the conductors are parallel; if they are at an angle to each other, then the result will be different.
The magnitude of B2 a radius d from wire 2 is B2 = m0I2/2pd. Using this in the expression for the force on wire 1 gives:
The direction of F1 is determined using the right hand rule -- actually, using both versions of the right hand rule, first to find the direction of B2, then use that to find the direction of F1. The force one wire 2 due to wire 1 is equal and opposite to F1, as required by Newton's third law. These forces tend to either repel or attract the wires from each other, depending on whether the currents are in the same, or in opposite directions, respectively.
Until recently, this force was used to define the ampere, the unit of current. The coulomb is then defined as the amount of charge passing a cross section of wire carrying 1A of current in 1 second.
Consider again the magnetic field generated by a straight wire, and imagine bending the wire into a circular loop. The center of the loop is the same distance from any small piece of wire, and the magnetic field generated by each little piece points in the same direction. The result is that the field at the center of the loop is enhanced.
If many loops of wire at stacked together, the field is enhanced even more. A stack of loops can be made by coiling the wire around a cylinder. Such a configuration is called a solenoid. A good solenoid is made with many windings of closely spaced wire. The magnetic field of a good solenoid is particularly simple. To good approximation, the field is uniform inside the solenoid, directed along the axis of the cylindrical shape, and outside the solenoid the field is approximately zero (well, much smaller than inside). The magnitude of the field inside the solenoid is:
An electron moves at a speed of 1.0×104m/s in a circular path of radius 2.0cm inside a solenoid. The magnetic field of the solenoid is perpendicular to the plane of the electron's path. Find (a) the strength of the magnetic field inside the solenoid and (b) the current in the solenoid if it has 25 turns per centimeter.
(a) To find the strength of the magnetic field, use the result for the radius of the circular path of a charged particle in a uniform magnetic field: r = m v / q B.
Use this to determine
B = m v / q r = (9.11×10-31kg)(1.0×104m/s) / (1.6×10-19C)(0.020m) = 2.8×10-6T.
(b) Now use the expression for the magnetic field in a solenoid to find the current.
We need the number of turns of wire per meter, n = 25/cm = 2500/m.
Since B = m0 n I,
I = B / m0 n = (2.8×10-6T) / (4p×10-7Tm/A)(2500/m) = 8.9×10-4A = 0.89mA.