Question

The terminals of a 0.700 V watch battery are connected by a 120 - m - long gold wire with a diameter of 0.200 mm.

Answer 1

Concepts and reason

The concept required to solve this question is the Ohm’s law, resistance, length, and area relation, and the cross section area of the wire.

Initially, calculate the resistance of the wire using the relation of the resistance with the area and the length of the wire. Finally, find the current using the Ohm’s law.

Fundamentals

The relation between the resistance, resistivity, area and the length is,

$R = \rho \frac{l}{A}$

Here, R is the resistance, $\rho$ is the resistivity, l is the length of the conductor, and A is the area.

The expression of the current using the ohm’s law is,

$I = \frac{V}{R}$

Here, I is the current, V is the voltage, and R is the resistor.

The cross section area of the wire is,

$A = \pi {\left( {\frac{d}{2}} \right)^2}$

Here, d is the diameter.

Substitute 0.200 mm for d.

$\begin{array}{c}\\A = \pi {\left( {\frac{{0.200{\rm{ mm}}}}{2}\left( {\frac{{{{10}^{ - 3}}{\rm{ m}}}}{{1{\rm{ mm}}}}} \right)} \right)^2}\\\\ = 3.14 \times {10^{ - 8}}{\rm{ }}{{\rm{m}}^2}\\\end{array}$

The relation between the resistance, resistivity, area and the length is,

$R = \rho \frac{l}{A}$

Substitute $3.14 \times {10^{ - 8}}{\rm{ }}{{\rm{m}}^2}$ for A, 120 m for l, and $2.44 \times {10^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}$ for $\rho$ .

$\begin{array}{c}\\R = \left( {2.44 \times {{10}^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}} \right)\frac{{120{\rm{ m}}}}{{\left( {3.14 \times {{10}^{ - 8}}{\rm{ }}{{\rm{m}}^2}} \right)}}\\\\ = 93.25{\rm{ }}\Omega \\\end{array}$

The expression of the current is,

$I = \frac{V}{R}$

Substitute 0.700 V for V and $93.25{\rm{ }}\Omega$ for R.

$\begin{array}{c}\\I = \frac{{0.700{\rm{ V}}}}{{93.25{\rm{ }}\Omega }}\\\\ = 7.50 \times {10^{ - 3}}{\rm{ A}}\left( {\frac{{1{\rm{ mA}}}}{{{{10}^{ - 3}}{\rm{ A}}}}} \right)\\\\{\rm{ = }}7.50{\rm{ mA}}\\\end{array}$

Ans:

The current in the wire is 7.50 mA.