Two long wires hang vertically. Wire 1 carries an upward current of 1.40 A. Wire 2, 20.0 cm to the right of wire 1, carries a downward current of 4.80 A. A third wire, wire 3, is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force. (a) Is this situation possible? Yes No Is it possible in more than one way? Yes No (b) Describe the position of wire 3. distance Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. cm direction (c) Describe the magnitude and direction of the current in wire 3. magnitude A direction

a. Yes, the solution is possible.

See Explanation below

b.

Wire 3 is located at the left of wire 1

c.

I3 = 1.2624A

magnitude = 4.8 * 10^-6N/m

Direction is down

Explanation:

b.

Given

I1 = 1A

I2 = 4.8A

L = 20.0cm = 0.2m

Calculating the force on wire 1

Let F = Force per unit length on wire 1 by wire 2 to the left

F = μo*I1*I2/2πL

Where the permeability of free space, μ0, is a physical constant used often in electromagnetism.

It is defined to have the exact value of 4π x 10-7 N/A2

F = 4π x 10-7 * 1 * 4.8/2π * 0.20

F = 2 * 10^-7 * 1 * 4.8/ 0.2

F = 4.8 * 10^-6N/m

So, the force on wire is 4.8 * 10^-6N/m and it is at the right.

This translates to; the magnitude of the force on wire 3

The magnitude = 4.8 * 10^-6N/m

Also, because the I2 > I1

Wire 3 must be closer to wire 1 than 2

We can conclude that Wire 3 is located at the left of wire 1

c.

Representing the distance between wore 1 and wore 3 with Y

Y + 0.2 = distance between wire 2 and 3

This gives the following expression using the formula in b) above

μo*I1*I3/2πY = μo*I3*I2/2π(Y + 0.20)

Divide through by μo * I3

I1/2πY = I2/2π(Y + 0.2)

Multiply both sides by 2π

I1/Y = I2/(Y + 0.2)

Cross Multiply

I1 * (Y + 0.2) = I2 * Y

Substitute the values of I1 and I2

1 * (Y + 0.2) = 4.8 * Y

Y + 0.2 = 4.8Y

4.8Y - Y = 0.2

3.8Y = 0.2

Y = 0.2/3.8

Y = 0.05263157894736842105263

Y = 0.0526m --- Approximated

Finally; μo*I1*I3/2πL = 4.8 * 10^-6N/m

I3 = 4.8x10^-6*2π * 0.0526 / (4π * 10^-7 *1)

I3 = 4.8x10^-6 * 0.0526 / (2* 10^-7 *1)

4π x 10-7 * 1 * I3 / 2π

I3 = 1.2624A

and its direction is down

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.