Question

Three coplanar parallel wires, each carrying a current of 10 A along the same direction, are placed with a separation 5.0 cm between the consecutive ones. Find the magnitude of the magnetic force per unit length acting on the wires. 

Answer 1

Let wires W₁, W₂ and W₃ be arranged as shown in the figure. 

Given:
Magnitude of current in each wire, i₁ = i₂ = i₃ = 10 A
The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by

\[\frac{F}{l} = \frac{\mu_0 i_1 i_2}{2\pi d}\]

So, for wire W₁,

 

\[\frac{F}{l} = \frac{F}{l}\text{ by wire }  W_2 + \frac{F}{l} \text{ by wire }  W_3 \]
\[ = \frac{\mu_0 \times 10 \times 10}{2\pi \times 5 \times {10}^{- 2}} + \frac{\mu_0 \times 10 \times 10}{2\pi \times 10 \times {10}^{- 2}}\]
\[ = \frac{2 \times {10}^{- 7} \times {10}^2}{5 \times {10}^{- 2}} + \frac{2 \times {10}^{- 7} \times {10}^2}{10 \times {10}^{- 2}}\]
\[ = 4 \times {10}^{- 4} + 2 \times {10}^{- 4} \]
\[ = 6 \times {10}^{- 4} N\]

For wire W₂,

 

\[\frac{F}{l} = \frac{F}{l} \text{ by wire } W_1 - \frac{F}{l} \text{ by wire } W_3 \]
\[ = \frac{\mu_0 \times 10 \times 10}{2\pi \times 5 \times {10}^{- 2}} - \frac{\mu_0 \times 10 \times 10}{2\pi \times 5 \times {10}^{- 2}} = 0\]

For wire W₃,

\[\frac{F}{l} = \frac{F}{l} \text{ by wire } W_1 + \frac{F}{l} \text{ by wire } W_2 \]
\[ = \frac{\mu_0 \times 10 \times 10}{2\pi \times 5 \times {10}^{- 2}} + \frac{\mu_0 \times 10 \times 10}{2\pi \times 10 \times {10}^{- 2}}\]

 = 6 × 10⁻⁴ N