The figure shows a copper strip of width W = 16.1 cm that has been bent to form

Question

The figure shows a copper strip of width W = 16.1 cm that has been bent to form a shape that consists of a tube of radius R = 4.4 cm plus two parallel flat extensions. Current i = 53 mA is distributed uniformly across the width so that the tube is effectively a one-turn solenoid. Assume that the magnetic field outside the tube is negligible and the field inside the tube is uniform. What are (a) the magnetic field magnitude inside the tube and (b) the inductance of the tube (excluding the flat extensions)?

Explanation / Answer

A. We imagine dividing the one-turn solenoid into N small circular loops placed along the width W of the copper strip. Each loop carries a current di = i/N.

Magnetic field inside solenoid is given by

B = u0*n*di = u0*(N/W)*(i/N)

B = u0*i/W

B = 4*3.14*10^-7*53*10^-3/0.161 = 4.13*10^-7 T

B. In solenoid

L = phi/i = B*A/i

L = pi*R^2*B/i

L = 3.14*0.044^2*4.13*10^-7/(53*10^-3)

L = 4.737*10^-8 H

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