0195136047.pdf

10 CIRCUIT CONCEPTS

EXAMPLE 1.1.3

Consider an infinitesimal length of 10−^6 m of wire whose center is located at the point (1, 0, 0),

carrying a current of2Ainthepositive direction ofx.

(a) Find the magnetic flux density due to the current element at the point (0, 2, 2).

(b) Let another current element (of length 10−^3 m) be located at the point (0, 2, 2), carrying

a current of1Ainthedirection of(− ̄ay+ ̄az). Evaluate the force on this current element

due to the other element located at (1, 0, 0).

Solution

(a)I 1 dl ̄ 1 = 2 × 10 −^6 a ̄x. The unit vectora ̄ 12 is given by

a ̄ 12 =

( 0 − 1 )a ̄x+( 2 − 0 )a ̄y+( 2 − 0 )a ̄z

√

12 + 22 + 22

=

(− ̄ax+ 2 a ̄y+ 2 a ̄z)

3

Using the Biot–Savart law, Equation (1.1.9), one gets

[B ̄ 1 ]( 0 , 2 , 2 )=

μ 0

4 π

I 1 d ̄l 1 × ̄a 12

R^2

whereμ 0 is the free-space permeability constant given in SI units as 4π× 10 −^7 H/m,

andR^2 in this case is{( 0 − 1 )^2 +( 2 − 0 )^2 +( 2 − 0 )^2 }, or 9. Hence,

[B ̄ 1 ]( 0 , 2 , 2 )=

4 π× 10 −^7

4 π

[

( 2 × 10 −^6 a ̄x)×(− ̄ax+ 2 a ̄y+ 2 a ̄z)

9 × 3

]

=

10 −^7

27

× 4 × 10 −^6 (a ̄z− ̄ay)Wb/m^2

= 0. 15 × 10 −^13 (a ̄z− ̄ay)T

(b)I 2 d ̄l 2 = 10 −^3 (− ̄ay+ ̄az)

dF ̄ 12 =I 2 dl ̄ 2 ×B ̄ 1

=

[

10 −^3 (− ̄ay+ ̄az)

]

×

[

0. 15 × 10 −^13 (a ̄z− ̄ay)

]

= 0

Note that the force is zero since the current elementI 2 dl ̄ 2 and the fieldB ̄ 1 due toI 1 dl ̄ 1

at (0, 2, 2) are in the same direction.

The Biot–Savart law can be extended to find the magnetic flux density due to a current-

carrying filamentary wire of any length and shape by dividing the wire into a number of

infinitesimal elements and using superposition. The net force experienced by a current loop can

be similarly evaluated by superposition.

Electric Potential and Voltage

When electrical forces act on a particle, it will possess potentialenergy. In order to describe the

potential energy that a particle will have at a pointx, theelectric potentialat pointxis defined as