A 10-cm-long nichrome wire is connected across the terminals of a 1.5 V battery.What is the current density inside the wire?

Let J be the Current Density.
First off, you must find the electric field strength E. This is V/m (potential over length in meters), so E = 1.5/.1
After this, you can find the current density J.
J = (omega) * E, omega being the conductivity. The conductivity of nichrome is 6.7*10^5, the unit being inverse ohm-meters.
Therefore J = 6.7*10^5 * 15 V/m^2

J = V/rho*L
rho can be 1e-6 to 1.5e-6 Ω-m
I’ll use 1.5e-6 and L = 0.1
J = 1.5/1.5e-6*0.1 = 1e7 C/sm²

Nichrome resistivity ρ nc = 1.00 ohm*mm^2/m
current I = V/R = V*s/(ρ nc*ℓ)
current density = I/s = V*s/(ρ nc*ℓ*s) = V/(ρ nc*ℓ) = 1.5/(1.00*0.1) = 15 A/mm^2

Answer Prime

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