Question

19. A$15$-cm-long nichrome wire is connected across the terminals of a $1.5\mathrm{V}$ battery.
a. What is the electric field inside the wire?
b. What is the current density inside the wire?
c. If the current in the wire is $2.0\mathrm{A}$, what is the wire's diameter?

Short answer

(a) The electric field inside the wire is $10V/m$.

(b) The current density inside the wire is $6.7\times {10}^6\mathrm{A}/{\mathrm{m}}^2$.

(c) The wire's diameter when the current in the wire $2.0A$is $6.2\times {10}^{-4}\mathrm{m}$

Step-by-step solution

Given information Part (a)

A $15$-cm-long nichrome wire is connected across the terminals of a $1.5\mathrm{V}$ battery with the electric field inside the wire.

Explanation Part (a)

The field in the wire is,

$E=\frac{\Delta V}{L}$

$=\frac{1.5V}{0.15m}$

$=10V/m$

Final answer Part (a)

The electric field inside the wire is $10V/m$.

Given information Part (b)

A $15$-cm-long nichrome wire is connected across the terminals of a $1.5\mathrm{V}$ battery with the current density inside the wire.

Explanation Part (b)

To find the current density$J$by using the electric field $E=10V/m$,
$J=\sigma E$

$J=6.7\times {10}^5\times 10V/m$

$=6.7\times {10}^6\mathrm{A}/{\mathrm{m}}^2$

Final answer Part (b)

The current density inside the wire is $6.7\times {10}^6\mathrm{A}/{\mathrm{m}}^2$.

Given information Part (c)

A$15$-cm-long nichrome wire is connected across the terminals of a $1.5\mathrm{V}$battery with the current in the wire is $2.0\mathrm{A}$.

Explanation Part (c)

Since the current density is $J=\frac{I}{A},$
where $A$ is the cross-section area. So,
$J=\frac{I}{\pi r^2}=\frac{4I}{\pi d^2}$

$\Rightarrow d=\sqrt{\frac{4I}{\pi J}}$
$d=\sqrt{\frac{4\times 2}{\pi \times 6.7\times {10}^6}}$

$=6.2\times {10}^{-4}\mathrm{m}$

Final answer Part (c)

The wire's diameter when the current in the wire $2.0A$is $6.2\times {10}^{-4}\mathrm{m}$.