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Difficulty: MediumMagnetic Force and Electromagnetism

Two long, parallel straight wires separated by a distance of 0.10 m0.10\text{ m} carry currents of 5.0 A5.0\text{ A} and 8.0 A8.0\text{ A} in opposite directions. What is the magnitude and nature of the magnetic force per unit length exerted between the two wires? (Take μ0=4π×10−7 T m A−1\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1})

  1. 8.0×10−5 N m−18.0 \times 10^{-5}\text{ N m}^{-1}, repulsiveAnswer
  2. B
    8.0×10−5 N m−18.0 \times 10^{-5}\text{ N m}^{-1}, attractive
  3. C
    4.0×10−5 N m−14.0 \times 10^{-5}\text{ N m}^{-1}, repulsive
  4. D
    4.0×10−5 N m−14.0 \times 10^{-5}\text{ N m}^{-1}, attractive

Answer

The magnetic force per unit length is 8.0×10−5 N m−18.0 \times 10^{-5}\text{ N m}^{-1} and the nature of the force is repulsive.
The magnetic force per unit length between two parallel conductors carrying currents I1I_1 and I2I_2 separated by distance dd is given by FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}. Substituting I1=5.0 AI_1 = 5.0\text{ A}, I2=8.0 AI_2 = 8.0\text{ A}, d=0.10 md = 0.10\text{ m}, and μ0=4π×10−7 T m A−1\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1} yields FL=8.0×10−5 N m−1\frac{F}{L} = 8.0 \times 10^{-5}\text{ N m}^{-1}. Furthermore, conductors carrying currents in opposite directions exert a repulsive force on each other.

Step-by-Step Solution

1
Identify the given parameters and formula for force per unit length between two parallel conductors.
Given I1=5.0 AI_1 = 5.0\text{ A}, I2=8.0 AI_2 = 8.0\text{ A}, d=0.10 md = 0.10\text{ m}, and μ0=4π×10−7 T m A−1\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}. Formula: FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}.
The magnetic field produced by one current-carrying conductor exerts a force on the second parallel conductor.
2
Substitute the values into the formula and evaluate the magnitude of the force per unit length.
\(\frac{F}{L} = \frac{(4\pi \times 10^{-7}) \times 5.0 \times 8.0}{2\pi \times 0.10} = \frac{2 \times 10^{-7} \times 40.0}{0.10} = 8.0 \times 10^{-5}\text{ N m}^{-1}\).
Simplifying 4π2π=2\frac{4\pi}{2\pi} = 2 allows direct scalar computation.
3
Determine the direction/nature of the magnetic force based on current direction.
Since the currents flow in opposite directions, the magnetic forces between the conductors are repulsive.
By the right-hand grip rule and magnetic force laws, parallel currents in opposite directions experience mutual repulsion.

Key Concept

Magnetic force per unit length between parallel current-carrying conductors
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Topics

Magnetic Force and Electromagnetism

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