The figure shows a copper strip of width W = 17.4 cm that has been bent to form

Question

The figure shows a copper strip of width W = 17.4 cm that has been bent to form a shape that consists of a tube of radius R = 3.4 cm plus two parallel flat extensions. Current I = 49 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

part a:

magnetic field inside a solenoid=mu*(N/L)*I

where mu=magnetic permeability=4*pi*10^(-7)

N=number of turns=1

L=length =17.4 cm=0.174 m

I=current=49 mA=0.049 A

then magnetic field=4*pi*10^(-7)*(1/0.174)*0.049=3.5388*10^(-7) T

part b:

inductance=N^2*mu*A/L

where A=cross section area=pi*R^2

so inductance=1^2*4*pi*10^(-7)*pi*0.034^2/0.174=2.6228*10^(-8) H

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