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

\[\frac{F_m}{l} = \frac{\mu_0 i_1 i_2}{2\pi d}\]   (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,

\[\frac{F_m}{l} = \frac{W}{l}\]
\[ \Rightarrow \frac{\mu_0 i_1 i_2}{2\pi d} = \lambda g\]

\[\Rightarrow \frac{2 \times {10}^{- 7} \times 50 \times i_2}{5 \times {10}^{- 3}} = 1 \times {10}^{- 4} \times 9 . 8\]
\[ \Rightarrow i_2 = \frac{9 . 8 \times {10}^{- 7}}{20 \times {10}^{- 7}} = 0 . 49\] A