SOLUTION Set the sum of the forces equal to zero and substitute the appropriate

Question

SOLUTION

Set the sum of the forces equal to zero and substitute the appropriate expressions. Notice that the magnetic force between the wires is repulsive.

The currents are equal, so I1 = I2 = I. Make these substitutions and solve for I2:

Substitute given values, finding I2, then take the square root. Notice that the weight per unit length, mg/ is given.

I = 7.07 A

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Remarks Another exercise showed that using Earth's magnetic field to levitate a wire required extremely large currents. Currents in wires can create much stronger magnetic fields than Earth's magnetic field in regions near the wire.

Question Why can't cars be constructed that can magnetically levitate in Earth's magnetic field? (Select all that apply.)

1

The current within the car must flow in a complete circuit, about half of which would exert forces against levitation.Magnetic forces are always very weak.The current needed would be excessive.Earth's magnetic field is in the wrong direction.

PRACTICE IT

EXERCISE HINTS:  GETTING STARTED | I'M STUCK!

Use the values from PRACTICE IT to help you work this exercise. If the current in each wire is doubled, how far apart should the wires be placed if the magnitudes of the gravitational and magnetic forces on the upper wire are to be equal?
d = ??? m

Explanation / Answer

magnetic force per length

F/L = uoi1i2/2pid

so here since i1 = i2

we have i^2 = F* 2pid/L * uo

i^2 = 1 e -4 * 2*3.14 * 0.1/(4*3.14 e -7)

i^2 = 50

i = each current = 7.07 Amps

Magnetic forces are always very weak.

The current needed would be excessive.

magnetic force per length

F/L = uoi1i2/2pid

so here since i1 = i2

we have i^2 = F* 2pid/L * uo

i^2 = 1 e -4 * 2*3.14 * 0.14/(4*3.14 e -7)

i^2 = 70

i = each current = 8.366 Amps

if i1 = i2 = 2i

then

F/L = 4 times of uoi1i2/2pid

so i^2 /d = consatnt

i2^2 /i1^2 = d2/d1

d2 = 4 d1

d2 = 0.56 m

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