A straight horizontal copper wire carries a current i = 30 A. The linear mass density of the wire is 45 g/m. What is the magnitude of the magnetic field needed to balance its weight?

Explanation

(1) Because `vec(L)` is directed horizontally (and the current is taken to be positive), Eq. and the right-hand rule for cross products tell us that `vec(B)` must be horizontal and rightward to give the required upward `vec(F)_(B)`.
The magnitude of `vec(F)_(B)` is `F_(B) = i LB sin phi` Because we want `vec(F)_(B)` to balance `vec(F)_(g)`, we want
`iLB sin phi = mg`,
where mg is the magnitude of `vec(F)_(g)` and m is the mass of the wire. We also want the minimal field magnitude B for `vec(F)_(B)` to balance `vec(F)_(g)`. Thus, we need to maximze `sin phi` in

To do so, we set `phi = 90^(@)`, thereby arranging for `vec(B)` to be perpendicular to the wire. We then have `sin phi = 1`, so yields
We write the result this way because we know m/L, the linear density of the wire. Substituting known data then gives us
`B = ((46.6 xx 10^(-3) kg//m)(9.8 m//s^(2)))/(28 A)`
`= 1.6 xx 10^(-2)T`
This is about 160 times the strength of Earth.s magnetic field.