Q4. What should the common current be in two parallel wires (16 cm long, 14 g mass, separated by 1.5 mm) so that the upper wire 'floats' above the lower wire with no tension in its suspension leads?

Background

Topic: Magnetic Fields and Forces Between Parallel Currents

This question tests your understanding of the magnetic force between two parallel wires carrying currents in opposite directions, and how this force can counteract gravity to achieve magnetic suspension.

Key Terms and Formulas

• Magnetic Force per Unit Length Between Parallel Wires:

• Total Magnetic Force:

• Gravitational Force:

• Where:

  • = permeability of free space ( T·m/A)

  • = currents in the wires (equal in this case)

  • = separation between wires (in meters)

  • = length of the wire (in meters)

  • = mass of the wire (in kilograms)

  • = acceleration due to gravity ( m/s)

Step-by-Step Guidance

1. Convert all quantities to SI units: m, kg, m.

2. Write the expression for the magnetic force per unit length between the wires: (since ).

3. Calculate the total magnetic force acting on the 16-cm wire: .

4. Set the total magnetic force equal to the gravitational force to achieve suspension: .

5. Substitute the formulas and solve for the current (do not compute the final value yet): .

Try solving on your own before revealing the answer!