Question

g The current through a 10 m long wire has a current density of 4 cross times 10 to the power of 6 space open parentheses bevelled A over m squared close parentheses. The wire conductivity is 2 cross times 10 to the power of 7 space open parentheses bevelled S over m close parentheses. Find the voltage drop across the wire. (Answer with the numeric value, don't write the unit V)

Answer 1

The value is
`V = 2 V`

Step-by-step explanation:

From the question we are told that

The length of the wire is
`l = 10 \ m`

The current density is
`J = 4*10^6 \ A/m^2`

The conductivity is
`\sigma = 2*10^(7) \ S/m`

Generally conductivity is mathematically represented as

`\sigma = (l)/(RA)`

Here R is the resistance which is mathematically represented as

`R = (V)/(I)`

Here I is the current which is mathematically represented as

`I = J * A`

So

`R = (V)/( J * A)`

And

`\sigma = (l)/((V)/( J * A) * A)`

=>
`\sigma = (l)/((V)/(J))`

=>
`V = (l * J)/(\sigma )`

=>
`V = (10 * 4*10^6)/(2*10^(7) )`

=>
`V = 2 V`