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U/J = fU /JdVol = brr PO;', ) 21Ch I'dI' = Po/ ' h In (~)<br />

v lls1C r 41C a<br />

Since U H = Po [ ' lln(~ ) =.!:.. LJ' , the inductance is<br />

41C a 2<br />

L = Poh In(~)<br />

21C a<br />

c) Calculate the inductance of tlus long inductor by usmg the formula<br />

(I) = LJ = f B· dA and your results for the magnetic fi eld in (a). To do this you<br />

ope" .mrfacl!<br />

must choose an appropriate open surface over whi ch to evaluate the magnetic flux. Does<br />

yo ur result calculated in this way agree with your result in (b)?<br />

cil' /<br />

/<br />

[ a<br />

I<br />

~<br />

~ ./<br />

The magnetic field is perpendicular to a rectangular<br />

surface shown in the figure. The magnetic flux tlu'ough a<br />

thin strip of area dA = ldr is<br />

,<br />

- -- --<br />

r -,<br />

-<br />

B- ...<br />

, ,<br />

i- - - - -<br />

-====:::<br />

~-... ::::~,<br />

"<br />

-' -- -,<br />

'~<br />

, , , ,<br />

~ - - -"" , -<br />

....... .. ..<br />

"-'0-- --------<br />

I<br />

Thus, the total magnetic flux is<br />

Thus, the inductance is<br />

whi ch agrees with that obtained in (b).<br />

~tJel/<br />

'M -{)<br />

- 90 fo<br />

d

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