Suppose you have a magnetic field directed in the $-\hat{z}$-direction, into the page. There is a flexible, circular loop situated on the page, in the $xy$-plane. You stretch it out in the $\pm x$-direction like a rubber band to change its area. Then you rotate it $90^\text{o}$ in the $xy$-plane. Finally, you rotate the loop $60^\text{o}$ in the $yz$-plane. What happens to the magnetic flux through the loop during these steps?

• The magnetic field is directed into the page.
• The steps for changing and rotating the loop are outlined in the problem statement.

• A description of the magnetic flux.

• The magnetic field is constant in time, and the same everywhere.
• The steps for changing and rotating the loop happen in a reasonable enough amount of time to describe the flux as the motions are happening.

• We represent magnetic flux through an area as

$$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A}$$

• We represent the steps with the following visual:

Since the magnetic field has a uniform direction, and the area of the loop is flat (meaning $\text{d}\vec{A}$ does not change direction either), then we can simplify the dot product: $$\vec{B} \bullet \text{d}\vec{A} = B\text{d}A\cos\theta$$

Since $B$ and $\theta$ do not change for different little pieces ($\text{d}A$) of the area, we can pull them outside the integral:

$$\int B\text{d}A\cos\theta =B\cos\theta \int \text{d}A = BA\cos\theta$$

Area for a square is just $A = L^2$, and $\theta$ is different for each loop:

\[ \Phi_B = \begin{cases} BL^2\cos 0 = 1.5 \cdot 10^{-4} \text{ Tm}^2 & \text{Loop 1} \\ BL^2\cos 90^\text{o} = 0 & \text{Loop 2} \\ BL^2\cos 42^\text{o} = 1.1 \cdot 10^{-4} \text{ Tm}^2 & \text{Loop 3} \end{cases} \]

Notice that we could've given answers for Loops 1 and 2 pretty quickly, since they are parallel and perpendicular to the magnetic field, respectively, which both simplify the flux calculation greatly.