Fig. 14 shows a straight solid conductor in which the steady current I is flowing upwards in the direction indicated. Suppose that the field strength at point A distant r meters from the centre of the conductor is H. Then, it means that if a unit N-pole is placed at A, it will experience a force of H newtons. The direction of this force would be tangential to the circular line of force passing thought A, If this unit N-pole is moved once round the conductor against this force, then work done is
= force × distance
= H × 2πr joules.
According to work law (It states that the net work in joules done on or by a unit N-pole in moving around any single completed path in a magnetic field is numerically equal to the ampere-turns linked with its path), this must be equal to the ampere-turns included within the path
Fig. 14. Magnetic field produced by current in a straight solid conductor.
Ampere-turns = NI
= 1 × I = I
Example 2. A long straight conductor carries steady current of 20 A. Find:
i. The intensity of magnetic field produced at a point 300 mm from the axis of the conductor.
ii. Flux density of field at that point.
Solution. Current carried by the conductor, I = 20 A
Distance, r = 300 mm = 0.3 m
(i) Magnetic field intensity, H :