Figure 2962 shows a cross section across a diameter of a lon

Figure 29-62 shows a cross section across a diameter of a long cylindrical conductor of radius a = 2.00 cm carrying uniform current 170 A. what is the magnitude of the current’s magnetic field at radial distance (a) 0, (b) 1.00 cm, (c) 2.00 cm (wire’s surface), and (d) 4.00 cm?

Solution

J = i / pi a²

= 170 / pi * 0.02² = 1.35 * 10⁵ A/m²

B = mu₀ J a² / 2 r

r = 1.00 cm

B = (4 pi * 10-7 * 1.35 * 10⁵ * 0.02² ) / 2 * 0.01

= 3.39 * 10-3 T

------------------------------------------------

r = 2.00 cm

B = (4 pi * 10-7 * 1.35 * 10⁵ * 0.02² ) / 2 * 0.02

= 1.69 * 10-3 T

------------------------------------------------

r = 4.00 cm

B = (4 pi * 10-7 * 1.35 * 10⁵ * 0.02² ) / 2 * 0.04

= 8.48 * 10-4 T