A square filamentary differential current loop, dL on a side, is centered at the origin in the
Question:
(a) Assuming that r >> dL, and following a method similar to that in Section 4.7, show that
(b) Show that
The square loop is one form of a magnetic dipole.
Transcribed Image Text:
Дol (dL)? sin@ 4лr2 dA = a, I(dL)? dH = (2 cos e a, + sin 0 ag) %3D 4лг3
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a Assuming that r dL and following a method similar to that in Sec 47 show that We begin with the ex...View the full answer
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