Question

18.5 a) using the data in table 18.1, compute the resistance of a copper wire 3 mm (0.12 in.) in diameter and 2 m (78.7 in.) long. (b) what would be the current flow if the potential drop across the ends of the wire is 0.05 v? (c) what is the current density? (d) what is the magnitude of the electric field across the ends of the wire?

Answer 1

Main Answer:

a) The resistance of the copper wire is approximately 0.115 ohms.

b) The current flow through the wire, given a potential drop of 0.05 V, is around 0.435 A.

c) The current density in the wire is roughly 1.58 x 10^6 A/m².

d) The magnitude of the electric field across the ends of the wire is about 0.026 V/m.

Therefore, the correct answer is All of them.

Step-by-step explanation:

In part (a), the resistance of the copper wire is calculated using the formula R = ρ * (L/A), where ρ is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area. The resistivity of copper is a fundamental property, and the cross-sectional area is determined from the diameter of the wire.

Moving on to part (b), Ohm's Law (V = IR) is applied to find the current flow through the wire. The potential drop (V) is given, and rearranging the formula to solve for current (I) provides the answer.

Part (c) involves calculating current density (J), which is the current per unit area. Current density is obtained by dividing the current by the cross-sectional area of the wire, providing an indication of how densely the electric current is flowing through the material.

Finally, in part (d), the magnitude of the electric field is determined using the formula E = V/L, where V is the potential drop and L is the length of the wire. This reflects the electric field strength across the wire.

These calculations help understand the electrical properties of the copper wire under the given conditions, shedding light on its resistance, current flow, current density, and electric field strength.

Therefore, the correct answer is All of them.

Answer 2

To compute the resistance of the copper wire, use the formula R = ρL/A where ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area. To find the current flow, use Ohm's Law: I = V/R where I is the current and V is the potential drop across the wire. The current density is given by J = I/A and the magnitude of the electric field can be found using E = V/L.

Part (a)

To compute the resistance of a copper wire, we can use the formula:

R = ρL/A

where R is the resistance, ρ (rho) is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area of the wire.

The resistivity of copper is approximately 1.7 × 10-8 Ωm. The length of the wire is 2 m and the diameter of the wire is 3 mm. To find the cross-sectional area, we can use the formula for the area of a circle:

A = πr2

where r is the radius of the wire. You can convert the diameter to radius by dividing it by 2. Once you have the radius, substitute the values into the formula for resistance to calculate the resistance of the wire.

Part (b)

To calculate the current flow, we can use Ohm's Law:

I = V/R

where I is the current, V is the potential drop across the wire, and R is the resistance of the wire. Substitute the given values to find the current flow.

Part (c)

The current density is given by the formula:

J = I/A

where J is the current density, I is the current flow, and A is the cross-sectional area of the wire. Substitute the values to find the current density.

Part (d)

The magnitude of the electric field across the ends of the wire can be found using the formula:

E = V/L

where E is the electric field, V is the potential drop across the wire, and L is the length of the wire. Substitute the given values to find the magnitude of the electric field.