Question

A square loop of wire with side length a carries a current I₁. The center of the loop is located a distance d from an infinite wire carrying a current I₂. The infinite wire and loop are in the same plane; two sides of the square loop are parallel to the wire and two are perpendicular as shown. (Figure 1)

Part A

What is the magnitude, F, of the net force on the loop?

Express the force in terms of I₁, I₂, a, d, and μ₀.

Part B

The magnetic moment \(\vec{m}\) of a current loop is defined as the vector whose magnitude equals the area of the loop times the magnitude of the current flowing in it m=IA, and whose direction is perpendicular to the plane in which the current flows. Find the magnitude, F, of the force on the loop from Part A in terms of the magnitude of its magnetic moment.

Answer 1

Part A

force between to parallel wires, \(F=\frac{\mu_{0} I_{1} I_{2} l}{2 \pi r}\)

here \(l\) is the length and \(r\) is the distance.

force acting on acting on the side nearer to the wire,

$$ F_{1}=\frac{\mu_{0} I_{1} I_{2} a}{2 \pi\left(d-\frac{a}{2}\right)} $$

this force is attractive

force acting on acting on the side farther to the wire,

$$ F_{2}=\frac{\mu_{0} I_{1} I_{2} \alpha}{2 \pi\left(d+\frac{a}{2}\right)} $$

this force is repul sive.

net force, \(F=F_{1}-F_{2}\)

$$ \begin{aligned} &=\frac{\mu_{0} I_{1} I_{2} a}{2 \pi\left(a-\frac{a}{2}\right)}-\frac{\mu_{0} I_{1} I_{2} a}{2 \pi\left(d+\frac{a}{2}\right)} \\ &=\frac{\mu_{0} I_{1} I_{2} a}{\pi}\left(\frac{1}{2 d-a}-\frac{1}{2 d+a}\right) \\ &=\frac{2 \mu_{0} I_{1} I_{2} a^{2}}{\pi\left(4 d^{2}-a^{2}\right)} \end{aligned} $$

Part B

magnetic moment, \(m=I_{1} a^{2}\)

therefore, \(F=\frac{2 \mu_{0} I_{2} m}{\pi\left(4 d^{2}-a^{2}\right)}\)

Answer 2

Answer 3