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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

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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

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