Question

A cross section of three parallel wires each carrying a current of 20 A. The current in wire B is out of the paper, while that in wires A and C are into the paper. If the distance R = 15.0 mm,

what is the magnitude of the force on a 200 cm length of wire C?

Answer 1

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The magnitude of force is
`F_R} = 5.33 \ m N`

Step-by-step explanation:

From the question we are told that

The current in wire A , B and C are
`I _a = I_b =I_c = I= 20 A`

The distance is
`R = 15.0mm = (15)/(1000) = 0.015m`

The length of wire C is
`L_c = 200cm = (200)/(100) = 2 m`

Generally the force exerted per unit length that is acting in between two current carrying conductors can be mathematically represented as

`(F)/(L) = (\mu_o )/(2 \pi) \cdot (I_1 I_2 )/(R)`

Where
`\mu_o` is the permeability of free space with a constant value of

`\mu_o = 4 \pi * 10^(-7) N/A^2`

When the current in the wire are of the same direction the force is positive and when they are in opposite direction the force is negative

Considering force between A and C

`F_(A -C) = (\mu_o )/(2 \pi) \cdot (I_a I_c )/(2R) * L`

Considering force between B and C

`F_(B -C) = (\mu_o )/(2 \pi) \cdot (I_b I_c )/(R) * L`

The resultant force is

`F_R = F_(B -C) - F_(A -C) = (\mu_o )/(2 \pi) \cdot (I_a I_c )/(2R) * L - (\mu_o )/(2 \pi) \cdot (I_b I_c )/(R) * L`

`F_R} = (\mu_o )/(2 \pi) \cdot (I * I )/(R) * L * ({1 - (1)/(2) })`

Substituting values

`F_R} = (4 \pi * 10^(-7))/(2 * 3.142) \cdot ( 20*20 )/(0.015) * 2 ({(1)/(2) })`

`F_R} = 5.33 *10^(-3) N`

`F_R} = 5.33 \ m N`