Magnetic Flux

Setup: A uniform magnetic field \(B = 0.50\text{ T}\) passes through a flat rectangular surface of area \(A = 0.20\text{ m}^2\). The angle between \(\textbf{B}\) and the surface normal is \(\phi = 60°\). Find \(\Phi_B\).

Because both \(B\) and \(\phi\) are constant over the entire flat surface, they factor out of the integral:

The remaining integral \(\int dA\) simply sums up all the tiny area patches, which gives the total area \(A\):

This is the flat-surface, uniform-field shortcut. You only pull out the integral when things are constant — if either \(B\) or \(\phi\) varied across the surface, you'd have to keep the integral.

This is the geometric projection factor — it tells you that only half of the field's magnitude is contributing to the flux. The other half is sliding along the surface and contributes nothing.

Setup: Same problem as Example 1 — \(B = 0.50\text{ T}\), \(A = 0.20\text{ m}^2\), \(\phi = 60°\) — but this time we use Form 2 to build different geometric intuition.

The magnetic field \(\textbf{B}\) can be split into two components relative to the surface:

Plug in numbers:

Since \(B_\perp\) is constant over the surface:

Setup: The same physical scenario, but now the field is given in component form:

The surface lies in the \(xz\)-plane, so its outward normal points in the \(\hat{j}\) direction. Therefore:

For a flat surface with uniform field:

Expand the dot product term by term:

Now use the fundamental dot product results for unit vectors:

So the \(\hat{i}\) term vanishes completely — that's the part of the field running parallel to the surface, which contributes no flux:

Setup: \(B = 0.80\text{ T}\), \(A = 0.30\text{ m}^2\). The magnetic field runs parallel to the surface — it is not piercing through at all, just sliding along it. Find \(\Phi_B\).

If \(\textbf{B}\) is parallel to the surface, it is perpendicular to the surface normal. Therefore the angle between \(\textbf{B}\) and the normal is:

Setup: A magnetic field varies with position:

The field points in the \(+\hat{i}\) direction everywhere. The surface is a rectangle lying in the \(yz\)-plane, with its normal also in the \(+\hat{i}\) direction. It spans \(x = 0\) to \(x = 3.0\text{ m}\) in one direction and has height \(L = 2.0\text{ m}\) in the \(z\)-direction.

Since \(B\) depends only on \(x\), we slice the rectangle into thin vertical strips, each at position \(x\), each with width \(dx\) and height \(L = 2.0\text{ m}\). The area of one strip is:

The field \(\textbf{B}\) points in \(+\hat{i}\) and the area normal also points in \(+\hat{i}\). They are parallel, so:

The field punches straight through — no projection penalty.

The flux contribution from one strip at position \(x\) is:

Integrate over the full width of the surface:

Factor out the constant \(4\):

Use the power rule \(\int x\,dx = \frac{x^2}{2}\):

Evaluate at the limits:

Setup: A uniform magnetic field is given in full 3D component form:

The surface is a flat rectangle of area \(A = 0.40\text{ m}^2\) lying in a tilted plane. The plane's outward unit normal is:

This normal is a unit vector (we'll verify that) pointing at 45° between the \(+y\) and \(+z\) axes. Find \(\Phi_B\).

Always check that \(|\hat{n}| = 1\) before using it. If it's not normalized, the area element is wrong.

For a flat surface with uniform field, the entire area contributes uniformly, so:

Now execute \(\Phi_B = \textbf{B}\cdot\textbf{A}\), expanding component by component:

Distribute. Only matching unit vectors survive (\(\hat{i}\cdot\hat{j}=0\), \(\hat{i}\cdot\hat{k}=0\), \(\hat{j}\cdot\hat{k}=0\), \(\hat{j}\cdot\hat{j}=1\), \(\hat{k}\cdot\hat{k}=1\)):

Setup: A magnetic field depends on position in the \(xy\)-plane:

The surface is a flat rectangle lying in the \(xy\)-plane, with its outward normal pointing in the \(+\hat{k}\) direction. It spans \(0 \le x \le 2\text{ m}\) and \(0 \le y \le 1\text{ m}\). Find \(\Phi_B\).

The surface lies in the \(xy\)-plane with normal \(\hat{k}\). The area element is:

Here we write \(dA = dx\,dy\) because we're integrating over a rectangle — tiny rectangular patches of width \(dx\) and height \(dy\).

Distribute the dot product through each term:

The first two terms vanish. The integrand collapses to just the constant \(2\).

Inner integral (over \(x\), holding \(y\) fixed):

Outer integral (over \(y\)):

Setup: A uniform magnetic field points in the \(+z\)-direction:

The surface is the curved wall of a cylinder of radius \(R\) and height \(h\), oriented with its axis along \(+z\). Find the flux through this curved lateral surface.

This one requires building \(d\textbf{A}\) from scratch using a parameterization.

A cylinder of radius \(R\) can be parameterized using two parameters: \(\phi \in [0, 2\pi)\) (angle around the axis) and \(z \in [0, h]\) (height up the axis). The position vector of any point on the curved wall is:

This traces out every point on the cylinder as \(\phi\) and \(z\) vary over their ranges.

To build \(d\textbf{A}\), we need the tangent vectors to the surface in the \(\phi\) and \(z\) directions.

Partial derivative with respect to \(\phi\) (tangent in the "around" direction):

Partial derivative with respect to \(z\) (tangent in the "up" direction):

These two tangent vectors lie in the surface. Their cross product gives a vector perpendicular to the surface — the outward normal — and its magnitude gives the area scaling factor.

Evaluate the \(3\times 3\) determinant:

Expand along the top row:

This is the outward normal vector field of the cylinder! It points in the radial direction \(\hat{r}\) and has magnitude \(R\):

So the vector area element is:

The magnitude \(R\,d\phi\,dz\) is the area of a tiny rectangular patch on the cylinder (arc length \(R\,d\phi\) times height \(dz\)). Makes sense.

Every term vanishes. The integrand is identically zero everywhere on the surface.