Two long parallel conductors are separated by a distance of 0.07 m. The currents are $I_{1} = 12, A$ and $I_{2} = 5, A$ , flowing in opposite directions. What is the magnitude of the force per unit length exerted on each conductor?

medium 3 marks Class 12 · Physics · Moving Charges and Magnetism

Answer

1.71 × 10⁻⁴ N m⁻¹

The correct answer is 1.71 × 10⁻⁴ N m⁻¹ — Hence the force per unit length is approximately $1.71 \times 1 0^{- 4}, N, m^{- 1}$ .

Step-by-step solution

For two long parallel conductors the force per unit length is $F / l = \frac{m u _{0}}{2 p i} \frac{I _{1} I _{2}}{d}$ .
Substitute the given values: $μ_{0} = 4 π \times 1 0^{- 7}, T, m / A$ ; $I_{1} = 12, A$ ; $I_{2} = 5, A$ ; $d = 0.07, m$ .
Compute the constant: $\frac{μ _{0}}{2 π} = \frac{4 π \times 1 0 ^{- 7}}{2 π} = 2 \times 1 0^{- 7}, T, m / A$ .
Then $F / l = 2 \times 1 0^{- 7} \times \frac{12 \times 5}{0.07} = 2 \times 1 0^{- 7} \times \frac{60}{0.07}$ .
Calculate $\frac{60}{0.07} \approx 857.14$ ; multiply by $2 \times 1 0^{- 7}$ gives $1.714 \times 1 0^{- 4}, N, m^{- 1}$ .
Hence the force per unit length is approximately $1.71 \times 1 0^{- 4}, N, m^{- 1}$ .

Common mistake: Using the wrong constant in the formula (e.g., forgetting the

1/2 p i

factor)

Concept tested: Force between parallel currents