A straight round copper conductor of radius `R = 5.0 mm` carries a current `I = 50 A` Find the potentail difference between the axis of the conductor and its surface. The concentration of the conduction electrons in copper is equal to `n = 0.9.10^(33) cm^(-3)`.

A straight round copper conductor of radius `R = 5.0 mm` carries a cur

1.	A straight round copper conductor of radius `R = 5.0 mm` carries a current `I = 50 A` Find the potentail difference between the axis of the conductor and its surface. The concentration of the conduction electrons in copper is equal to `n = 0.9.10^(33) cm^(-3)`.
Answer» The electrons in the conductor are drifting with a speed of, `v_(d) = (J)/("ne") = (I)/(pi R^(2) "ne")`, where `e` = magnitude of the charge on the electron, `n` = concentration of the conductin electrons. The magentic field inside the conductor due to this current is given by, `B_(varphi) (2pi r) = pi r^(2) (I)/(pi R^(2)) mu_(0)` or, `B_(varphi) = (mu_(0))/(2pi) (I r)/(R^(2))` A radial electric field `vB_(varphi)` must come into being in equilibrium. Its `P.D.` is, `Delta varphi = int_(0)^(R) (I)/(pi R^(2) n e) (mu_(0))/(2pi) (Ir)/(R^(2)) dr = (I)/(pi R^(2)n e) ((mu_(0))/(4pi) I) = (mu_(0) I^(2))/(4pi R^(2) n e)`

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A straight round copper conductor of radius `R = 5.0 mm` carries a current `I = 50 A` Find the potentail difference between the axis of the conductor and its surface. The concentration of the conduction electrons in copper is equal to `n = 0.9.10^(33) cm^(-3)`.

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