Question

A long, hollow, cylindrical conductor (with inner radius 2.0 mm and outer radius 4.0 mm) carries a current of 24 A distributed uniformly across its cross section. A long thin wire that is coaxial with the cylinder carries a current of 24 A in the opposite direction. What is the magnitude of the magnetic field (a) 1.0 mm, (b) 3.0 mm, and (c) 5.0 mm from the central axis of the wire and cylinder

Answer 1

Answer:

a) 0.0048 T

b) 0.00093 T

c) 0 T

Explanation:

B = μ(0) [I - i'] / 2πr

where I = 24, and i' = 0, this means that

B = [12.568*10^-7 (24 - 0) ] / 2 * 3.142 * 0.001

B = 3.016*10^-5 / 0.006284

B = 0.0048 T

b), formula is the same, except for change in values, so we have

I = 24, and

i' = (3y2 - 2y2) / (4y2 - 2y2) * 24

i' = 10, and substituting these in the equation, we have

B = [12.568*10^-7 (24 - 10) ] / 2 * 3.142 * 0.003

B = (12.568*10^-7 * 14) / 0.018852

B = 1.76*10^-5 / 0.018852

B = 0.00093 T

and for the last one, we have

I = 24 and i' = 24 also, so that

B = [12.568*10^-7 (24 - 24) ] / 2 * 3.142 * 0.005

B = 0 / 0.03142

B = 0