Question

In: Physics

A silver wire with resistivity 1.59 x 10^-8 ? x m carries a current density of...

A silver wire with resistivity 1.59 x 10^-8 ? x m carries a current density of 4.0 mm^2. What is the magnitude of the electric field inside the wire?

A) 0.064 V/m

B) 2.5 V/m

C) 0.040 V/m

D) 0.10 V/m

Solutions

Expert Solution

The equation you need for this question is E = pJ where E is the electric field, p is the resistivity, and J is the current density. The one "trick" in this problem is with the units. Current density represents current per unit area and is typically given in units of amperes / m^2. Is it possible thta you may have mistyped something and that the current density should be in units of 4.0 A / mm^2? In that case, we can change this over to m^2 using dimentionsal analysis:

4.0 I / mm^2 * (1000 mm / 1 m) * (1000 mm / 1 m) = 4E6 I / m^2

Once we know that, all we have to do is plug the values into the equation:

E = pJ

E = 1.59E-8 ohms-m * 4E6 A / m^2 = .0636 V / m, which is answer choice A

*Note - the reason I asked about the units is because if you look at the plugged in equation, you'll see that the units line up after the conversions we did above. Solely looking at the units, the answer ends up in terms of ohms-meter-amperes per meters-squares. One of the meters cancels out from both the top and the bottom, leaving ohms-amperes per meter. We know from the equation V = IR that ohms times amperes equals volts, giving us final answer units of V / m

Hope this helps!


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