Institutt for fysikk, NTNU TFY4155/FY1003 Elektrisitet og magnetisme Vår 2007

Løsningsforslag til øving 10

Veiledning uke 11

Oppgave 1

The conductor may be divided into sylindracal tubes with inner radius r, outer radius r + dr, and hence a cross section with area

dA = 2πr dr

The current running in such a tube is

dI = j · dA = j_{0} e^{−r/R} · 2πr dr

Total current I is then found by integrating dI over the cross section of the conductor, i.e., by letting r vary from 0 to R:

I = ∫ dI
= ∫_{0}^{R} j_{0} e^{−r/R} · 2πr dr
= 2πj_{0} R^{2} ∫_{0}^{1} x e^{−x} dx
= 2πj_{0} R^{2} (1 − \frac{2}{e})

Here, we have substituted x = r/R so that dr = R dx and r = R x. The integral is easily solved with integration by parts.