Question bank — Magnetic flux Φ = ∫B·dA

True. Magnetic field lines form closed loops with no start or end (no monopoles), so $\oint \mathbf{B}\cdot d\mathbf{A}=0$ for any closed surface.

False. If the loop is edge-on ($\theta =90^{\circ }$), $\cos \theta =0$ and flux is zero no matter how strong $\mathbf{B}$ is — orientation matters as much as strength.

False. The dot product collapses the two vectors into a single number, so $\Phi$ is a scalar (it has magnitude and sign, but no direction).

False. It is the angle between $\mathbf{B}$ and the surface normal $\hat{\mathbf{n}}$. "Normal's the boss."

False. The flux $\Phi$ through each single turn is unchanged; what doubles is the flux linkage $N\Phi$, which is what appears in Faraday's EMF.

True. If $\theta >90^{\circ }$ the field enters the "back" of the surface, $\cos \theta <0$, so $\Phi <0$ — the sign encodes which way lines pierce relative to the chosen normal.

True. Flipping $\hat{\mathbf{n}}$ replaces $\theta$ by $180^{\circ }-\theta$, so $\cos \theta \to -\cos \theta$; the magnitude of flux is a physical fact, the sign is a bookkeeping convention.

True. Only the projection of the area onto the plane perpendicular to $\mathbf{B}$ catches lines; $\Phi =B(A\cos \theta )$.

Error: "perpendicular to the field" describes the loop's plane; the normal is then parallel to $\mathbf{B}$, so $\theta =0^{\circ }$ and $\Phi =BA$ (maximum), not zero.

Error: $\Phi =BA$ only holds when the field is also perpendicular to a flat surface. In general you still need $BA\cos \theta$, and a curved surface or non-uniform field forces the integral.

Error: there is no magnetic charge. The magnetic Gauss's law is $\oint \mathbf{B}\cdot d\mathbf{A}=0$ always — the right side is fixed at zero, unlike the electric case.

Error: lines lying in the plane skim along the surface and pierce nothing. Here $\theta =90^{\circ }$ (normal $\perp \mathbf{B}$), giving $\Phi =0$.

Error: flux is field times area, so its unit is the weber ($1\text{ Wb}=1\text{ T}\cdot {\text{m}}^2$). Tesla is the unit of $\mathbf{B}$ alone.

Error (avoidable work): flux through the curved cap equals flux through its flat circular rim, because in a uniform field $\oint \mathbf{B}\cdot d\mathbf{A}=0$ over the closed surface (cap + disc). Use the easy flat disc.

Error: only if both catch the same field orientation. A large loop held edge-on has zero flux while a small face-on loop has positive flux.

Because only the part of the field along the normal pierces the surface; the dot product $\mathbf{B}\cdot d\mathbf{A}=BA\cos \theta$ automatically extracts that perpendicular component and discards the sliding-along part.

Because magnetic field lines are unbroken closed loops — every line that goes in comes back out. There is no monopole inside to act as a source or sink, so the ins and outs cancel exactly.

So that over each patch the field is essentially constant and the simple rule $\mathbf{B}\cdot d\mathbf{A}$ applies; summing and taking the limit (the integral) rebuilds the true total for varying fields or curved surfaces.

Because induced EMF responds to the change in how many lines thread the loop — that count is the flux. A changing $\mathbf{B}$, a changing area, or a changing tilt can all drive it, and flux packages all three.

The normal is the one direction that measures how "face-on" the patch is to the field; projecting the field onto it gives the true line-piercing contribution, which is exactly what flux should count.

Tilting shrinks the projected area $A\cos \theta$ that faces the field, so fewer lines pass through; at $\theta =90^{\circ }$ the projection vanishes and flux hits zero.

Zero. $\cos 90^{\circ }=0$, so $\Phi =BA\cos 90^{\circ }=0$ — the lines run parallel to the surface and none pierce it.

Zero. With $\mathbf{B}=0$, the integrand $\mathbf{B}\cdot d\mathbf{A}=0$ at every patch regardless of orientation, so $\Phi =0$.

Zero. With $dA$ summing to nothing there is no area for lines to cross; $\Phi =0$ trivially.

$\cos \theta$ becomes slightly negative, so $\Phi$ crosses smoothly from positive to negative as the loop tips past edge-on — the field now enters through the chosen "back" of the surface.

Exactly zero. In a uniform field every line entering one side exits the other; more fundamentally $\oint \mathbf{B}\cdot d\mathbf{A}=0$ holds for all closed surfaces.

The full maximum, $100\%$. Here $\theta =0$, $\cos 0=1$, so $\Phi =BA$ — every line pierces head-on.

It starts at the maximum $BA$ (rate of change zero there) and falls to $0$ at edge-on; because $\Phi \propto \cos \theta$, the flux changes fastest near $\theta =90^{\circ }$ and slowest near $\theta =0^{\circ }$.