Question

Three long wires all lie in an xy plane parallel to the x axis. They intersect the y axis at the origin, y = d, and y = 2d, where d = 10 cm. The two outer wires each carry a current of 6.8 A in the positive x direction.

1. What is the magnitude of the force on a 2.8 m section of either of the outer wires if the current in the center wire is 3.3 A.

(a) in the positive x direction and (b) in the negative x direction?

Answer 1

`F = 2.55*10^(-4) \ N`

F =
`3.808 * 10^(-6) \ N`

Step-by-step explanation:

The magnetic field due to the first wire can be written as:

`B_1 = (\mu _o I_1)/(2 \pi (2 d))`

`B_1 = (\mu _o I_1)/(4 \pi d)`

The magnetic field due to the second wire is as follows:

`B_2 = (\mu _o I_2)/(2 \pi d)`

The net magnetic field B =
`B_1 +B_2`

`B = (\mu _o I_1)/(4 \pi d) +(\mu _o I_2)/(2 \pi d) \\ \\ B = ( \mu_o )/(2 \pi d)((I_1)/(2)+ I_2)`

Also; the magnetic force on wire segment l = 2.8 m

`F = I_1lB = ( \mu_o )/(2 \pi d)((I_1)/(2)+ I_2)`

Replacing all the values ; we have :

`F = ( 4 \pi * 10^(-7)*6.8*2.8)/(2 \pi * 10*10^(-2))((6.8)/(2)+ 3.3)`

`F = 2.55*10^(-4) \ N`

b)

Magnetic field due to the first wire is :

`B_1 = (\mu _o I_1)/(4 \pi d)`

`B_1 = (4.0*10^(-7)*6.8)/(4 \pi *10*10^(-2))`

`B_1 = 6.8*10^(-6) \ T`

Magnetic field due to the second wire is :

`B_2 = (\mu _o I_2)/(2 \pi d) \\ \\ B_2 = (4*10^(-7)*3.3)/(2 \pi *10*10^(-2))`

`B_2 = 6.6*10^(-6)T`

`B_1>B_2`

Net magnetic field B =
`B_1 - B_2`

B =
`(6.8*10^(-6) - 6.6*10^(-6))T`

B =
`2*10^(-7) \ T`

Magnetic force F =
`I_1lB`

F =
`6.6*2.8 *2*10^(-7)`

F =
`3.808 * 10^(-6) \ N`