Question

The figure shows a cross section of three parallel wires each carrying a current of 15 A. The currents in wires A and C are out of the paper, while that in wire B is into the paper. If the distance R = 5.0 mm, what is the magnitude of the force on a 4.0-m length of wire C?

Answer 1

Answer:

0.018 N

Explanation:

From the given information;

We understood that A and C possess the same current direction, thus the force between them is said to be an attractive force.

Also, between B and C, there exist different current directions, thus, the force between them is said to be a repulsive force.

So, the force per unit length magnitude are as follows:

$\dfrac{Fa}{L} = \dfrac{\mu_oI_1I_2}{2 \pi d}$

$\dfrac{Fa}{L} = \dfrac{(4 \pi \times 10^{-7})\times (15)^2}{2 \pi (0.01)}$

$\dfrac{Fa}{L} = 0.0045 N$

Similarly;

$\dfrac{Fb}{L} = \dfrac{\mu_oI_1I_2}{2 \pi d}$

$\dfrac{Fb}{L} = \dfrac{(4 \pi \times 10^{-7})\times (15)^2}{2 \pi (0.005)}$

$\dfrac{Fb}{L} =0.009\ N$

Thus, the net force acting on 4.0 - m length of the wire C is:

$F_{net} =\bigg( \dfrac{Fb}{L} - \dfrac{Fa}{L} \bigg) \times 4.0$

$F_{net} =\bigg( 0.009-0.0045 \bigg) \times 4.0$

$F_{net} =\bigg( 0.0045 \bigg) \times 4.0$

$\mathbf{F_{net} = 0.018 \ N}$

Thus, the magnitude of the force on a 4.0-m length of wire C = 0.018 N