Lesson 20 (1) Magnetic field of a current carrying straight wire A segment of a straight wire lies on the z-­‐axis occupying the length between the coordinates z1 and z2 and carries a current I in the positive z direction. The magnetic field it creates at a point P depends on the distance between P and the z-­‐
axis. Let the shortest distance be r . Without loss of generality, we can take the point P to lie on the x-­‐y plane. A small segment of length dz at the point Z with z-­‐
coordinate z is at a distance of r 2 + z 2 from P. The field due to this current element has magnitude µ I dz
dB = 0 2 2 sin " 4! r + z
where ! is the angle between ZP and the z-­‐axis. This field is in the direction of the fingers of the right hand when the thumb is lined up with the current. The field due to all segments are in the same direction. From the diagram, we see that sin ! = sin (" ! ! ) =
r
r2 + z2
The magnitude of the field due to the whole segment is therefore z
µ 0 Ir 2
dz
B=
!
2
4! z1 ( r + z 2 )3 2
With the substitution z = r tan ! where ! is the angle between PZ and PO as shown, we find "
µI 2
µ I sin "2 " sin "1
B = 0 ! cos " d " = 0
4! r "1
2! r
2
The definitions of !1 and !2 are as shown, from which we find 1 sin !1 =
z1
2
2
1
r +z
sin !2 =
z2
2
r + z22
so that the angles can be negative. In the limit z1 ! "# z2 ! +# , we have !1 = ! " 2 !2 = + " 2 so that the magnetic field due to an infinitely long current wire is µI
B = 0 2! r
Example: Find the magnetic field at the point P due to the bent wire with dimensions shown. Solution: Magnetic field due to the infinitely long segments to the left and right of P are both zero. For the magnetic field due to the segment ab, !1 = 0 !2 = " 4 Bab =
µ 0 I sin ! 4 ! sin 0 µ 0 I 2 µ 0 I 2
=
=
2! L
2
2! L 4
!L 8
int o the paper For the other two segments bc and cd, µ I sin ! 4 ! sin (! ! 2 ) µ 0 I 2 µ 0 I 2
Bbc = 0
=
=
2! L
2
2! L 2
!L 4
µ I sin(0) ! sin(! ! 4) µ 0 I 2 µ 0 I 2
Bcd = 0
=
=
2! L
2
2! L 8
!L 8
2 int o the paper int o the paper µ0 I 2
int o the paper The net field is Bab + Bbc + Bvd = ! L 2
(2) Magnetic field due to multiple infinitely long parallel current wires The fields due to the individual wires should be added vectorially to obtain the actual field, as the following example shows. Example: As shown, two parallel infinitely long straight wires carry the same current of 10A in the same direction (into the paper), and are 1m apart. Find the magnetic field at the locations P, and Q, which are both 0.2m from the left wire, but on opposite sides of the left wire. Solution: The circular field lines that go through the point P from the left and right wires are shown. They help to determine the directions of the magnetic fields at the point P due to these wires, which are shown as arrows, and have magnitudes BL and BR. We have µ I 4π × 10 −7 × 10
µ I 4π × 10 −7 × 10
BL = 0 =
= 10 −5 T
BR = 0 =
= 0.25 × 10 −5 T 2πr
2π × 0.2
2πr
2π × 0.8
The resultant field at P has magnitude BP = BL − BR = 0.75 × 10 −5 T and points downward. Similarly, for the point Q, we have BL = 10 −5 T upward ;
BR =
4π × 10 −7 × 10
= 0.17 × 10 −5 T upward 2π × 1.2
so that BQ = BL + BR = 1.17 × 10 −5 T
upward 3 (3) Force between parallel current wires When we have two current carrying wires, each wire finds itself in the magnetic field created by the other, and so will experience a force. The simplest case is that of two infinitely long wires parallel to each other. Suppose the currents are in the same direction, being into the paper as shown. Let I 1 and I 2 be the magnitude of the currents, and r be the distance between the wire. The magnetic field due to the left wire at the location of the right wire is downward as shown, and its magnitude is µ I
B = 0 1 2r
The force on a length L of the right wire due to this field is also as shown and its magnitude is µ II
F = I 2 LB = 0 1 2 L 2r
Thus the right wire is attracted to the left wire. Similarly it can be shown that the left wire is also attracted to the right wire by an equal force. If the direction of the current in the right wire is reversed, the direction of the force on it is also reversed. The wire is now repelled by the left wire. Likewise, the left wire is also repelled by the right wire. Therefore we have the conclusion that parallel currents attract, anti-­‐parallel currents repel, which is just the opposite of what happens to the electrostatic forces between like and unlike charges. Definition of Coulomb The formula we have derived for the force between currents forms the basis of defining the ampere, the unit of current, because we have chosen (for convenience that becomes apparent in an more advanced discussion) µ 0 = 4π × 10 −7 . If we make the substitutions I1 = I 2 = 1A, r = 1m, L = 1m, we find F = 2π × 10 −7 N . So 1A is that current that causes such a force in the arrangement alluded to. Once we have defined the ampere, the unit of charge coulomb is defined to be that amount of charge flowing through a cross-­‐section of wire carrying 1A in one second. With the coulomb thus defined, the coulomb constant k , or the related quantity of permittivity of free space ε 0 , is then determined by experiments. 4