Question

What would the Hall voltage be if a 2.00-T field is applied across a 10-gauge copper wire (2.588 mm in diameter) carrying a 20.0-A current

Answer 1

The hall voltage is
`\epsilon =1.45 *10^(-6) \ V`

Step-by-step explanation:

From the question we are told that

The magnetic field is
`B = 2.00 \ T`

The diameter is
`d = 2.588 \ mm = 2.588 *10^(-3) \ m`

The current is
`I = 20 \ A`

The radius can be evaluated as

`r = (d)/(2)`

substituting values

`r = (2.588 * 10^(-3))/(2)`

`r = 1.294 *10^(-3) \ m`

The hall voltage is mathematically represented as

`\episilon = B * d * v_d`

where
`v_d` is the drift velocity of the electrons on the current carrying conductor which is mathematically evaluated as

`v_d = (I)/(n * A * q )`

Where n is the number of electron per cubic meter which for copper is

`n = 8.5*10^(28) \ electrons`

A is the cross - area of the wire which is mathematically represented as

`A = \pi r^2`

substituting values

`A = 3.142 * [ 1.294 *10^(-3)]^2`

`A = 5.2611 *10^(-6) \ m^2`

so the drift velocity is

`v_d = (20 )/( 8.5*10^(28) * 5.26 *10^(-6) * 1.60 *10^(-19) )`

`v_d = 2.7 *10^(-4 ) \ m/s`

Thus the hall voltage is

`\epsilon = 2.0 * 2.588*10^(-3) * 2.8 *10^(-4)`

`\epsilon =1.45 *10^(-6) \ V`