28.31.
Solve:
(a)
Let the new current be
I
′
such that
I
′
=
2
I
where
I
is the original current. The current density
is
JI
A
=
. Since the area of cross section of the wire remains the same, we have
A
I
J
I
J
==
′
′
⇒
′ =
′
=
J
I
I
JJ
2
That is, doubling the current doubles the current density.
(b)
Conductionelectron density is a material’s characteristic value, and does not depend on external factors. Thus
n
′
=
n
.
(c)
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This note was uploaded on 03/14/2009 for the course PHY PHYA22 taught by Professor Abbas during the Spring '09 term at University of Toronto.
 Spring '09
 Abbas
 Physics, Current

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