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Solution: Click For PDF Version (a)
Electric
field E, the
resistivity
and the
resistance R of a
conducting wire of length l, cross-sectional area A,
and resistivity
Therefore,
where
Therefore, the electric field at the surface of the wire will be
(b)
We will use
the Ampere’s law for finding the magnetic field at the surface
of the wire. If the radius of the wire is r,
the magnetic field B
and the current
Therefore,
(c)
As the
electric field
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)
carries a current of
25.0 A. We have to calculate
(a) the electric field,
(b) the magnetic field,
and (c) the
Poynting vector magnitude for a point on the surface of the wire.
,
and the current density j
are related as
,
the current flowing through a conductor of cross-sectional area A
is
.
It is given that the resistance per unit length of the copper wire
.
and the magnetic field
are orthogonal at the surface of the current carrying wire, the
magnitude of the Poynting vector will be given by