0195136047.pdf

1.1 ELECTRICAL QUANTITIES 7

Insulatorsare materials that do not allow charge to move easily. Examples include glass,
plastic, ceramics, and rubber. Electric current cannot be made to flow through an insulator, since
a charge has great difficulty moving through it. One sees insulating (ordielectric) materials often
wrapped around the center conducting core of a wire.
Although the term resistance will be formally defined later, one can say qualitatively that
a conductor has a very low resistance to the flow of charge, whereas an insulator has a very
high resistance to the flow of charge. Charge-conducting abilities of various materials vary in
a wide range.Semiconductorsfall in the middle between conductors and insulators, and have
a moderate resistance to the flow of charge. Examples include silicon, germanium, and gallium
arsenide.

Current and Magnetic Force

The rate of movement of net positive charge per unit of time through a cross section of a conductor
is known ascurrent,

i(t)=

dq

dt

(1.1.5)

The SI unit of current is the ampere (A), which represents 1 coulomb per second. In most
metallic conductors, such as copper wires, current is exclusively the movement of free electrons
in the wire. Since electrons are negative, and since the direction designated for the current
is that of the net positive charge movement, the charges in the wire are thus moving in the
direction opposite to the direction of the current designation. The net charge transferred at a
particular time is the net area under the current–time curve from the beginning of time to the
present,

q(t)=

∫t

−∞

i(τ) dτ (1.1.6)

While Coulomb’s law has to do with the electric force associated with two charged bodies,
Ampere’s law of forceis concerned with magnetic forces associated with two loops of wire carrying
currents by virtue of the motion of charges in the loops. Note that isolated current elements do
not exist without sources and sinks of charges at their ends; magnetic monopoles do not exist.
Figure 1.1.2 shows two loops of wire in freespace carrying currentsI 1 andI 2.
Considering a differential elementdl ̄ 1 of loop 1 and a differential elementdl ̄ 2 of loop 2,
the differential magnetic forcesdF ̄ 21 anddF ̄ 12 experienced by the differential current elements
I 1 dl ̄ 1 , andI 2 dl ̄ 2 , due toI 2 andI 1 , respectively, are given by

dF ̄ 21 =I 1 d ̄l 1 ×

(

μ 0

4 π

I 2 d ̄l 2 × ̄a 21

R^2

)

(1.1.7a)

dF ̄ 12 =I 2 dl ̄ 2 ×

(

μ 0

4 π

I 1 d ̄l 1 × ̄a 12

R^2

)

(1.1.7b)

wherea ̄ 21 anda ̄ 12 are unit vectors along the line joining the two current elements,Ris the distance
between the centers of the elements,μ 0 is the permeability of free space with units of N/A^2 or
commonly known as henrys per meter (H/m). Equation (1.1.7) reveals the following:

1. The magnitude of the force is proportional to the product of the two currents and the
   product of the lengths of the two current elements.