of Sources Magnetic fields
Outline for Today Ch 30, sec 30.2 - 30.3
Magnetic force between two parallel conductors Ampere's Law
Magnetic Force Between Two Parallel Conductors
Since a current carrying conductor has with it an associated magnetic field, two such conductors parallel to each other will exert magnetic forces on one another.
fig. 30.08
We can use these forces to derive the unit of current (Ampere). A note on directions:
We work out the direction of the magnetic field, B2
We work out the direction of the magnetic force of, say, the the magnetic field set up by wire 1 on wire 2...
30.2 - 1
Magnetic Force Between Two Parallel Conductors II
We want to find the magnetic force exerted on one wire by the magnetic field of the other. To begin, consider the magnetic force on wire 1 due to wire 2:
fig. 30.08
F = IL B
F1 = Il B2
F1 = I1 lB2
lB
F1 = I1 lB2 = I1 l
o I2 2a
=
o I1 I2 l 2a
If we were to do the same calculation for the other wire, we'd get the same thing. So, we can write that, in general, the force on one wire due to the other is:
F1 o I1 I2 = l 2a
The force between two parallel wires
30.2 - 2
Magnetic Force Between Two Parallel Conductors II
The force between the two wires can be used to define the ampere
Ampere: When the force per unit length between two long, parallel wires that -7 carry identical currents and are separated by 1m is 2 x 10 N/m, the current in each wire is defined to be 1 A.
With this in mind we can go back and define the Coulomb (the unit of charge):
Coulomb: When a conductor carries a steady current of 1 A, the quantity of charge that flows through a cross-section of the conductor in 1s is 1C.
30.2 - 3
Examples
1. In Figure P30.17, the current in the long, straight wire is I1 = 5.00 A and the wire lies in the plane of the rectangular loop, which carries the current I2 = 10.0 A. The dimensions are c = 0.100 m, a = 0.150 m, and = 0.450 m. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire. problem 30.17 in the text
fig. p30.65 fig. p30.17
2. An infinitely long straight wire carrying a current I1 is partially surrounded by a loop as shown in Figure P30.65. The loop has a length L, radius R, and carries a current I2. The axis of the loop coincides with the wire. Calculate force the exerted on the loop. problem 30.65 in the text
30.2 - 4
Andre Ampere
Ampere's Law
As we saw in the last day, a current carrying wire produces a surrounding magnetic field
fig. 30.09
1775-1838
note that when there is no current in the wire, the needles on the compass point north. when there is a current in the wire, the needles point in the direction of the magnetic field. If the direction of the current is reversed, then so is the direction that the needles point (consistent with the right-hand rule). The lines of magnetic field around the wire are circular. The intensity of the magnetic field decreases as you get further from the wire.
30.3 - 1
Ampere's Law II
To obtain an expression for Ampere's law, begin with the line integral
B ds
fig. 30.09b
We can take this line integral over a circle around the wire. In this case the magnetic field is constant at all points over the line integral. We can remove it from the integral.
B ds = B
ds
From the previous day, we know that
B=
o I 2r
also,
ds = 2r circumference
around the circle!
B ds = B
ds =
o I 2r = o I 2r
30.3 - 2
Ampere's Law III
Although the result was calculated for a closed circle around a wire, it holds for any closed loop (called an Amperian loop). Hence, Ampere's law can be expressed as:
B ds = o I
I is the total steady state current passing through any surface bound...