Question

A wire of length L has a rectangular cross-section with height H and width W. The wire has a hole all the way through it of area a. The wire is made of material with constant resistivity ρ1 . A battery with a known voltage is applied across the ends of the wire. a. Find the electric field inside the material and the current flowing through the wire. b. If the hole is totally filled with a material that has resistivity ρ2 , what will be the electric field everywhere inside the wire, and what current will flow through the wire?

Answer 1

Assuming the voltage applied accross the conductor to be V

A) Electric field inside a conductor is given by,

E*L = V (since E = ∆V/∆r, where ∆V is the potential difference and ∆r is the length)

Therefore, E = V/L.

Now the Resistance of the conductor is given by,

R = *L/A, (where is resistivity, L is length of conductor, and A is the area of the cross section)

Here, =₁, L = L , & A = H*W - a

Therefore R_{​​​​​​1} = ₁*L/(H*W-a)

Now current through a resistor is given by V/R. (Using ohm's law)

Current through the conductor = V/R_{​​​​​​1} = V*(H*W-a)/₁*L

B) in this case definition of electric field remains same, therefore the electric field inside the conductor remains same throughout,

That is, E = V/L

Current flowing through the conductor,

R_{​​​​​​1} =₁*L/(H*W-a). (as calculated above)

R_{​​​​​​2} = ₂*L/a

Now both the resistors are in parallel,

Therefore equivalent resistance will be,

1/R_{​​​​​​eq} = 1/R_{​​​​​​1} + 1/R_{​​​​​​2}

And current through the conductor is given by V/R_{​​​​​​eq}

V/R_{​​​​​​eq} = V*(1/R_{​​​​​​1} + 1/R_{​​​​​​2}) = V/R_{​​​​​​1} + V/R_{​​​​​​2}

I = V*(H*W-a)/(₁*L) + V*a/(₂*L)