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. We arbitrarily choose a counterclockwise direction for the current, and convenient coordinates axes.

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:

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

It remains to find the direction of the force, for which we will use the Right Hand Rule. You should be able to convince yourself based on the coordinates we have chosen that the force on the left side is in the $+\hat{z}$ direction, and the force on the right side is in the $-\hat{z}$ direction.

This means that the net force on the loop is $0$, the loop's center of mass won't move! However, the opposing forces on opposite sides will cause the loop to spin – there is a torque! The calculation for the torque is shown below, with a diagram included to show visually what happens.

The Loop Rotates

The calculation is here:

$$\vec{\tau} = \vec{r} \times \vec{F} = \vec{r}_\text{left} \times \vec{F}_\text{left} + \vec{r}_\text{right} \times \vec{F}_\text{right} = \left(-\frac{L}{2} \hat{x}\right) \times \left(IBL \hat{z}\right) + \left(\frac{L}{2} \hat{x}\right) \times \left(-IBL \hat{z}\right) = IBL^2 \hat{y}$$

  • 184_notes/examples/week12_force_loop_magnetic_field.1510107658.txt.gz
  • Last modified: 2017/11/08 02:20
  • by tallpaul