D.4 Induction (HL only)

Magnetic flux is given by $\Phi =BA\cos \theta$. Here $B=0.80\text{T}$, $A=0.05{\text{m}}^2$, and the field is vertical while the coil lies flat, meaning the field is perpendicular to the plane of the coil.

The angle $\theta$ in the flux formula is measured between the magnetic field and the normal to the coil surface. Since the coil is flat (horizontal) and the field is vertical, the field is parallel to the normal, so $\theta =0°$ and $\cos 0°=1$.

$\Phi =BA\cos \theta =0.80\times 0.05\times 1=0.040\text{Wb}$

The magnetic flux through the coil is $0.040\text{Wb}$.

We need the magnetic flux $\Phi =BA\cos \theta$ when a vertical field passes through a horizontal coil.

Zero flux would occur if $\theta =90°$, meaning the field is parallel to the plane of the coil. Here the field is perpendicular to the plane (parallel to the normal), so flux is maximum, not zero. Eliminate $0\text{Wb}$.

$0.080=0.80\times 0.10$, which would require an area of $0.10{\text{m}}^2$. The coil area is $0.05{\text{m}}^2$, not $0.10{\text{m}}^2$. Eliminate this option.

$0.056\approx 0.80\times 0.05\times \cos 45°$. There is no basis for $\theta =45°$ here since the field is aligned with the normal. Eliminate this option.

$\Phi =0.80\times 0.05\times \cos 0°=0.040\text{Wb}$ is the only consistent answer.

The problem states the plane of the loop makes $30°$ with the field. The angle $\theta$ in $\Phi =BA\cos \theta$ is between the field and the normal to the loop. If the plane makes $30°$ with $B$, then the normal makes $90°-30°=60°$ with $B$.

$\Phi =BA\cos \theta =2.0\times 6.0\times 10^{-3}\times \cos 60°$ $=2.0\times 6.0\times 10^{-3}\times 0.5=6.0\times 10^{-3}\text{Wb}$

The magnetic flux is $6.0\times 10^{-3}\text{Wb}$. The common error is using $\cos 30°$ instead of $\cos 60°$, which gives the wrong answer.

The angle given ($30°$) is between the plane and the field, but we need the angle between the normal and the field: ${\theta }_{\text{normal}}=90°-30°=60°$.

This equals $2.0\times 6.0\times 10^{-3}\times 1=1.2\times 10^{-2}$, corresponding to $\theta =0°$ (maximum flux). The field is not perpendicular to the plane, so this is incorrect.

This equals $2.0\times 6.0\times 10^{-3}\times \cos 30°\approx 1.04\times 10^{-2}$. This is the distractor that uses the given $30°$ directly as the angle from the normal — but $30°$ is measured from the plane, not the normal.

Zero flux requires $\theta =90°$ from the normal, i.e. the field is parallel to the plane. Since the plane makes only $30°$ with the field, this is not the case.

$\Phi =2.0\times 6.0\times 10^{-3}\times \cos 60°=6.0\times 10^{-3}\text{Wb}$ is correct.