# Mass needed to balance the magnetic force on upper rod

## Homework Statement

Two straight rods 60 cm long and 2.0 mm apart in a current balance carry currents of 18 A each in opposite directions. What mass must be placed on the upper rod to balance the magnetic force of repulsion?

$\mu_0 = 4 \pi * 10^-7 ~\frac {T * M} {A}$
$g = 9.81 ~m/s$

## Homework Equations

Magnetic force between parallel wires: $~{dF_{12}} = I_2{d\ell_2} \frac {\mu_0 I_1 } {2\pi R}$
Force due to gravity: $~F = mg$

## The Attempt at a Solution

My initial assumption (and the textbook's method) is to find the magnetic force and set it equal to F = mg. However, I keep getting the magnetic force as being $\frac {18^2*(.6)*(4 \pi * 10^-7)} {2 \pi *(.002)}$ = .01944 N, and when I set that equal to mg and divide by 9.81 I get a mass of 0.00198 kg = 1.98 g, which is not the correct answer. I know the correct answer is 0.99 g (the problem is multiple choice), but for the life of me I can't seem to figure out how to get to that answer.

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