Question

We will use the equation for the magnetic force between two parallel wires applied to sides 1 and 3 of the loop to find the net force resulting from these opposing force vectors. For the total magnetic force for sides 1 and 3 of the loop, we have the following. The expression above evaluates to or a total magnetic force with magnitude mu N directed toward the

Answer 1

First of all a figure must be attached with the question. The numerical values should be given clearly.

Here the force is given by,

"First of all a figure must be attached with the question. The numerical values should be given clearly. Here the force is given by, F^vector = F^vector 1 + F^vector 2 = mu_0 I_1 I_2 l/2pi (1/c + a - 1/c) i^^= mu_0 I_1 I_2 l/2pi (-a/c(c + a)I From the question, we can get the following information, I_1 I_2 = 87 A^2, l = 0.595 m, a = 0.15 m c(c + a) = 0.025 m^2 Using these values we get, F^vector = 4pi times 10^-7 times 87 times 0.595/2 pi times (- 0.15/0.025)i = -6.2 times 10^-5 i^^N Hence the magnitude of the total force is,

From the question, we can get the following information,

$I_1 I_2 = 87 \hspace{2mm} A^2, \hspace{8mm} l = 0.595 \hspace{2mm } m, \hspace{8mm} a = 0.15 \hspace{2mm} m$

$c(c+a) = 0.025 \hspace{2mm} m^2$

Using these values we get,

$\vec{F} = \frac{4\pi \times 10^{-7} \times 87 \times 0.595}{2\pi} \times \left( - \frac{0.15}{0.025}\right ) \hspace{2mm} \hat{i} = -6.2 \times 10^{-5} \hspace{2mm} \hat{i} \hspace{2mm} \text{N}$

Hence the magnitude of the total force is,

$|\vec{F}| = 6.2 \times 10^{-5} \hspace{2mm} N = 62 \hspace{2mm} \mu N$

And the force is directed towards the -x axis.