A long, straight wire is fixed horizontally and carries a current of 50.0 A. A second wire having linear mass `10^(-4) Kg` directly above this wire at a separation of 5.0 mm. What current should this second wire carry such that the magnetic repulsion can balance its weight?

Explanation

To solve the problem step by step, we need to find the current that the second wire should carry so that the magnetic repulsion force balances its weight.

### Step 1: Understand the Forces Acting on the Second Wire
The weight of the second wire acts downwards and is given by:
\[ W = mg \]
where \( m = 10^{-4} \, \text{kg} \) is the mass of the wire and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.
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