Two long, straight, parallel horizontal conductors are separated vertically by a distance … — JAMB UTME

Two long, straight, parallel horizontal conductors are separated vertically by a distance of $2.0\text{ cm}$ . The upper conductor has a mass per unit length of $0.04\text{ kg/m}$ and carries a steady current of $50\text{ A}$ . Assuming the currents in the two conductors flow in opposite directions so that the resulting magnetic force is repulsive, what current (in amperes) must flow through the lower conductor to magnetically levitate and balance the weight of the upper conductor? (Take $g = 9.8\text{ m/s}^2$ and $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$ )

Answer: 784 A

Answer

The required current in the lower conductor is

784\text{ A}

Equating magnetic repulsion per unit length

\frac{\mu_0 I_1 I_2}{2\pi d}

to weight per unit length

\lambda g

gives

\frac{(2 \times 10^{-7}) \times 50 \times I_2}{0.02} = 0.04 \times 9.8

, which simplifies to

5 \times 10^{-4} I_2 = 0.392

, yielding

I_2 = 784\text{ A}

Step-by-Step Solution

Equate the upward repulsive magnetic force per unit length to the downward gravitational weight per unit length.

\frac{\mu_0 I_1 I_2}{2\pi d} = \lambda g

For the upper conductor to levitate in vertical static equilibrium, the upward magnetic force per meter must exactly balance its weight per meter.

Substitute all given physical values in SI units into the force balance equation.

\frac{(4\pi \times 10^{-7}) \times 50 \times I_2}{2\pi \times 0.02} = 0.04 \times 9.8

Converting distance

d = 2.0\text{ cm} = 0.02\text{ m}

and using mass density

\lambda = 0.04\text{ kg/m}

sets up a single equation with unknown

I_2

Simplify both sides of the equation.

5 \times 10^{-4} I_2 = 0.392

Calculating

\frac{2 \times 10^{-7} \times 50}{0.02} = 5 \times 10^{-4}\text{ N/(A}\cdot\text{m)}

and

0.04 \times 9.8 = 0.392\text{ N/m}

Solve for the unknown current

I_2

I_2 = \frac{0.392}{5 \times 10^{-4}} = 784\text{ A}

Dividing the weight per meter by the magnetic force coefficient yields the exact required current magnitude.

Key Concept

Interaction force between parallel current-carrying conductors and mechanical equilibrium

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