Question

Learning Goal: To learn to apply the concept of current density and microscopic Ohm's law.

A slab of metal of volume V is made into a rod of length L . The rod carries current I when the electric field inside is E.

A)Find the resistivity of the metal P.

B)The rod is now stretched so that its length is doubled. If the electric field remains the same, what is the new current I' in the rod?

C)A piece of copper is made into a rod with a square cross-section. The side of the square is 2.00 centimeters. The resistivity of copper is 1.7 *10^-8 om*m. . An unknown electric field E, directed along the rod, creates a current of 12.0 amperes through the rod. Find the magnitude of E.

Answer 1

Concepts and reason

The concepts required to solve this question is Ohm’s law, resistivity, and current density.

Initially, write the expression of the resistance. Then, rearrange the expression for the resistivity. Write the resistance in terms of voltage and the current using Ohm’s law. Substitute this resistance value in resistivity expression.

Rearrange the expression of resistivity for current. Later, find the relation of current with the length and then find the current when the length gets doubled.

Finally, write the expression of the current density in terms of current and area and then write the current in terms of resistivity and the electric field. Then equate both equation and rearrange for electric field.

Fundamentals

The expression of the resistance is,

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

Here, R is the resistance, $\rho$ is the resistivity, L is the length, 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 resistance.

The expression of the potential difference in terms of electric field is,

$V = EL$

Here, V is the potential difference, E is the electric field, L is the length of the conductor.

The expression of the current density is,

$J = \frac{I}{A}$

Here, J is the current density, I is the current, and A is the area.

The current density in terms of electric field and resistivity is,

$J = \frac{E}{\rho }$

(A)

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

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

Rearrange the expression for resistance.

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

The expression of the resistance is,

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

Rearrange the expression for resistivity.

$\rho = \frac{{RA}}{L}$

Substitute $\frac{V}{I}$ for R.

$\begin{array}{c}\\\rho = \left( {\frac{V}{I}} \right)\frac{A}{L}\\\\ = \frac{{VA}}{{IL}}\\\end{array}$

Substitute EL for V.

$\rho = \frac{{\left( {EL} \right)A}}{{IL}}$

Substitute V’ for LA.

$\rho = \frac{{EV'}}{{IL}}$

The resistivity of the material is.

$\rho = \frac{{EV'}}{{IL}}$

Rearrange the expression for the current.

$I = \frac{{EV'}}{{L\rho }}$

The current is inversely proportional to the length from the above expression.

$I \propto \frac{1}{L}$

If the length of the metal is doubled then the current is,

$I' \propto \frac{1}{{2L}}$

Take ratio of I and I’.

$\begin{array}{c}\\\frac{{I'}}{I} = \frac{L}{{2L}}\\\\I' = \frac{I}{2}\\\end{array}$

(C)

The expression for electric field is,

$E = \frac{{I\rho }}{A}$

Substitute $1.7 \times {10^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}$ for $\rho$ , 12.0 A for I, and 2.00 cm for A.

$\begin{array}{c}\\E = \frac{{\left( {12.0{\rm{ A}}} \right)\left( {1.7 \times {{10}^{ - 8}}{\rm{ }}\Omega \cdot {\rm{m}}} \right)}}{{\left( {2.00{\rm{ cm}}} \right){{\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{{\rm{ 1 cm}}}}} \right)}^2}}}\\\\ = 5.1 \times {10^{ - 4}}{\rm{ N/C}}\\\end{array}$

The electric field is $5.1 \times {10^{ - 4}}{\rm{ N/C}}$ .

Ans: Part A

The resistivity of the metal is $\frac{{EV'}}{{IL}}$ .

Part B

The new current in the rod is $\frac{I}{2}$ .

Part C

The magnitude of the electric field is $5.1 \times {10^{ - 4}}{\rm{ N/C}}$ .