A rectangular current loop is located near a long, straight wire that carries a current of 12 A. The current in the loop is 25 A. Determine the magnitude of the net magnetic force that acts on the loop.

Explanation: The net magnetic force acting on the loop is the superposition of the forces on each side of the loop. The forces on the two sides perpendicular to the long straight wire will cancel (right hand rule). The magnitude of the magnetic force between two wires with currents I1 = 12 A and I2 = 25 A of length L = 0.50 m is

\begin{displaymath}

F = \frac{\mu_o I_1 I_2 L}{2 \pi d}\end{displaymath}

so the forces on the near and far wires are

\begin{displaymath}

F_n = 2.7 \times 10^{-4} N\end{displaymath}

\begin{displaymath}

F_f = 1.2 \times 10^{-4} N\end{displaymath}

The force Fn is an attractive one, while Ff is a repulsive one (right hand rule). The magnitude of the force is therefore

F = Fn - Ff

\begin{displaymath}

F = 1.5 \times 10^{-4} N\end{displaymath}

answer: option c.

explanation: