Question

A long, straight wire is fixed horizontally and carries a current of 50.0 A. A second wire having linear mass density 1.0 × 10⁻⁴ kg m⁻¹ is placed parallel to and 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?

Answer 1

Given:
Magnitude of current, i₁ = 10 A
Separation between two wires, d = 5 mm
Linear mass density of the second wire, λ = 1.0 × 10⁻⁴ kgm⁻¹
Now,
Let i₂ be the current in the second wire in opposite direction.
Thus, the magnetic force per unit length on the wire due to a parallel current-carrying wire is given by
$\frac{F_{\mathrm{m}}}{l}=\frac{{\mathrm{\mu }}_0{i}_1{i}_2}{2\mathrm{\pi }d}$ (upwards)
Also,
Weight of the second wire, W = mg
Weight per unit length of the second wire, $\frac{W}{l}=\lambda g$ (downwards)

Now, according to the question,
$$\begin{gathered} \frac{F_{\mathrm{m}}}{l}=\frac{W}{l} \\ \Rightarrow \frac{{\mathrm{\mu }}_0{i}_1{i}_2}{2\mathrm{\pi }d}=\lambda \mathrm{g} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{2\times {10}^{-7}\times 50\times i_2}{5\times {10}^{-3}}=1\times {10}^{-4}\times 9.8 \\ \Rightarrow i_2=\frac{9.8\times {10}^{-7}}{20\times {10}^{-7}}=0.49\mathrm{A} \end{gathered}$$