184_notes:examples:week12_force_loop_magnetic_field

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Suppose you have a square loop (side length $L$) of current $I$ situated in a uniform magnetic field $\vec{B}$ so that the magnetic field is parallel to two sides of the loop. What is the magnetic force on the loop of current?

Facts

  • The loop is a square with side-length $L$.
  • The magnetic field is parallel to two sides of the loop, and has a magnitude $B$.
  • The current in the loop is $I$.

Lacking

  • The magnetic force on the loop.

Approximations & Assumptions

  • The current is in a steady state.
  • The magnetic field does not change.
  • There are no outside forces to consider.

Representations

  • We represent the magnitude of force on a current-carrying wire in a magnetic field as

$$\left| \vec{F} \right|=IBL\sin\theta$$

  • We represent the situation with the diagram below.

Square Loop in a B-field

In order to break down our approach into manageable chunks, we split up the loop into its four sides, and proceed. It is easy to find the magnitude of the force on each side, since $\theta$ for each side is just the angle between the magnetic field and the directed current.

Theta for Each Side

This gives the following magnitudes:

\[ \left| \vec{F} \right| = \begin{cases} IBL\sin 0 = 0 & \text{top} \\ IBL\sin 0 = 0 & \text{bottom} \\ IBL\sin \frac{\pi}{2} = IBL & \text{left} \\ IBL\sin \frac{\pi}{2} = IBL & \text{right} \\ \end{cases} \]

We can reason that the direction of the field is $+\hat{z}$ because of the Right Hand Rule. We also don't care about the magnetic field from Wire 2 at the location of Wire 2, since Wire 2 cannot exert a force on itself. Now, it remains to calculate the magnetic force.

Since we know the magnetic field, the next thing we wish to define is $\text{d}\vec{l}$. Wire 2 has current directed with $\hat{y}$ in our representation, so we can say $$\text{d}\vec{l} = \text{d}y \hat{y}$$

This gives $$\text{d}\vec{l} \times \vec{B}_1 = \frac{\mu_0 I_1}{2 \pi r}\text{d}y \hat{x}$$

Lastly, we need to choose the limits on our integral. We can select our origin conveniently so that the segment of interest extends from $y=0$ to $y=L$. Now, we write:

$$\vec{F}_{1 \rightarrow 2 \text{, L}} = \int_0^L I_2 \text{d}\vec{l} \times \vec{B}_1 = \int_0^L \frac{\mu_0 I_1 I_2}{2 \pi r}\text{d}y \hat{x} = \frac{\mu_0 I_1 I_2 L}{2 \pi r} \hat{x}$$

Force Per Length

Notice that this force is repellent. The equal and opposite force of Wire 2 upon Wire 1 would also be repellent. Had the two current been going in the same direction, one can imagine that the two wires would attract each other.

  • 184_notes/examples/week12_force_loop_magnetic_field.1510093991.txt.gz
  • Last modified: 2017/11/07 22:33
  • by tallpaul