How to Calculate Magnetic Flux When the Angle Is to the Coil’s Plane

A question says a magnetic field makes an angle of \(30^\circ\) with the plane of a coil. You remember the magnetic flux equation, \(\Phi_B=BA\cos\theta\), so you substitute \(30^\circ\).

The calculation looks reasonable, but you’ve used the wrong angle.

Here’s the important distinction: the magnetic flux equation uses the angle between the magnetic field and the coil’s normal, not the angle between the field and the coil’s plane.

If the angle \(\alpha\) is given to the plane, you can calculate the magnitude of magnetic flux directly using:

\[ \boxed{|\Phi_B|=BA\sin\alpha} \]

Let’s work through why this uses sine, how to recognise which angle an exam question gives you, and how to apply the method when a coil has multiple turns.

01The 90° difference: coil’s plane versus its normal

Imagine holding a rectangular sheet of cardboard in front of you. The flat surface is the plane. Now imagine an arrow sticking straight out from the centre of the cardboard.

That arrow represents the normal: a line perpendicular to the plane.

The area vector points along this normal, and its magnitude is equal to the area enclosed by the coil.

There are two angles to keep separate:

• \(\alpha\): the angle between the magnetic field and the plane of the coil.
• \(\theta\): the angle between the magnetic field and the normal to the coil.

Because the normal is perpendicular to the plane, these angles add to \(90^\circ\):

\[ \theta=90^\circ-\alpha \]

The standard magnetic flux equation is:

\[ \Phi_B=BA\cos\theta \]

Substituting the relationship between the angles:

\[ \Phi_B=BA\cos(90^\circ-\alpha) \]

Using the trigonometric identity \(\cos(90^\circ-\alpha)=\sin\alpha\):

\[ \boxed{|\Phi_B|=BA\sin\alpha} \]

That’s the entire adjustment. You don’t need a new physics equation. You just need to identify which angle the question has given you.

Why does sine make physical sense?

Think about the direction of the magnetic field.

If the field runs parallel to the coil’s surface, very little of it passes through the enclosed area. The magnetic flux is zero.

If the field points directly through the surface, the flux is at its maximum.

Only the component of the magnetic field perpendicular to the coil’s plane contributes to magnetic flux.

When the field makes an angle \(\alpha\) with the plane, that perpendicular component is:

\[ B_\perp=B\sin\alpha \]

Therefore:

\[ \Phi_B=B_\perp A \]

\[ \boxed{|\Phi_B|=BA\sin\alpha} \]

This component approach is often the easiest way to remember the equation. You’re calculating how much of the magnetic field actually passes through the coil, rather than how much runs along its surface.

How the angle changes the magnetic flux

The following table shows what happens as the field rotates relative to the coil.

Notice the pattern.

• Small angle to the plane means small magnetic flux.
• Large angle to the plane means large magnetic flux.
• A field perpendicular to the plane produces maximum flux.

This gives you a useful way to check your answer before reaching for your calculator.

If a question says the field is almost parallel to the coil’s plane, your calculated flux should be close to zero. If you get something close to \(BA\), you’ve probably used the wrong angle.

02The equation and its symbols

For a flat coil in a uniform magnetic field:

\[ \boxed{|\Phi_B|=BA\sin\alpha} \]

Each symbol has a specific meaning.

One weber (Wb) is equivalent to one tesla multiplied by one square metre:

\[ 1\,\mathrm{Wb}=1\,\mathrm{T\,m^2} \]

Remember that \(A\) is the area enclosed by the coil, not the length of wire used to make it.

What if the coil has multiple turns?

So far, we’ve calculated the magnetic flux through one turn.

A real coil can contain many turns, with each turn experiencing approximately the same magnetic flux. This introduces magnetic flux linkage.

Flux linkage is the magnetic flux through one turn multiplied by the number of turns:

\[ \boxed{\lambda=N\Phi_B} \]

where:

• \(\lambda\) is magnetic flux linkage, commonly expressed in Wb-turn.
• \(N\) is the number of turns.
• \(\Phi_B\) is the magnetic flux through one turn, in Wb.

Combining the two equations:

\[ \boxed{|\lambda|=NBA\sin\alpha} \]

This equation is useful when an HSC question asks for the total flux linkage rather than the flux through an individual turn.

Be careful: If the question asks for magnetic flux, don’t automatically multiply by \(N\). Only include the number of turns when calculating flux linkage.

The simple multiplication assumes that all turns link the same flux, as is normally assumed in these HSC coil problems.

05The common mistake: Two different 30° angles

Here’s a comparison worth remembering.

Suppose two identical coils are placed in the same uniform magnetic field.

• Coil A has a magnetic field making an angle of \(30^\circ\) with the plane.
• Coil B has a magnetic field making an angle of \(30^\circ\) with the normal.

Are their magnetic fluxes equal?

No. The two angles describe different physical orientations.

Let’s use the same values for both coils:

\[ B=0.40\,\mathrm{T} \]

\[ A=0.020\,\mathrm{m^2} \]

Coil A: 30° to the plane

The angle to the normal is:

\[ \theta=90^\circ-30^\circ=60^\circ \]

Therefore:

\[ \Phi_A=BA\cos60^\circ \]

\[ =0.40\times0.020\times0.5 \]

\[ \boxed{\Phi_A=0.0040\,\mathrm{Wb}} \]

Coil B: 30° to the normal

This time, the given angle is already the angle required by the cosine equation.

\[ \Phi_B=BA\cos30^\circ \]

\[ =0.40\times0.020\times0.8660 \]

\[ \boxed{\Phi_B=0.00693\,\mathrm{Wb}} \]

Coil B experiences more magnetic flux.

The difference comes entirely from orientation. The magnetic field, area, and coil shape are identical.

This is a common source of lost marks in HSC Physics. Students see the number \(30^\circ\), remember the cosine equation, and substitute without checking what the angle is measured from.

The number alone doesn’t tell you which trigonometric function to use. The physical reference line does.

06A reliable method for exam questions

When you see a magnetic flux question involving an angle, follow this sequence.

1. Identify the reference.

Does the question give the angle between the field and the coil’s plane, or between the field and the normal?

2. Convert the angle if necessary.

If the angle is given to the plane:

\[ \theta=90^\circ-\alpha \]

3. Identify what the question wants.

• Magnetic flux through one turn: use \(\Phi_B=BA\cos\theta\).
• Magnetic flux linkage: use \(\lambda=N\Phi_B\).

4. Calculate the area.

Check whether the coil is circular, square, or rectangular. Convert all dimensions into metres before calculating.

5. Check the physical result.

A field parallel to the plane should give zero flux. A field perpendicular to the plane should give maximum flux.

There is one more detail worth understanding before you tackle the questions: the sign of magnetic flux.

Does the angle to the plane tell you the sign?

Not by itself.

The area vector points perpendicular to the coil, but there are two possible directions for this normal. You choose one as the positive direction.

The signed magnetic flux is:

\[ \Phi_B=\vec B\cdot\vec A \]

or:

\[ \Phi_B=BA\cos\theta \]

where \(\theta\) is the angle between the magnetic field and the chosen area vector.

If the magnetic field has a component in the direction of the chosen normal, flux is positive. If it has a component in the opposite direction, flux is negative.

The angle between the field and the plane normally gives an acute angle between \(0^\circ\) and \(90^\circ\). This tells you the magnitude of the flux, but not necessarily its sign.

For most straightforward calculation questions, the magnitude is all you need. However, when a question involves a rotating coil, a changing magnetic field, or induced EMF, the sign can become important.

07Questions and solutions

Try these five questions in order. The first checks the basic substitution, while the later questions combine orientation, flux linkage, and changes in the area vector.

Question 1

A single-turn rectangular coil has an area of \(0.018\,\mathrm{m^2}\). It is placed in a uniform magnetic field of \(0.50\,\mathrm{T}\). The field makes an angle of \(30^\circ\) with the plane of the coil.

Calculate the magnitude of the magnetic flux through the coil.

Solution 1

The magnetic flux magnitude is:

\[ \boxed{4.5\times10^{-3}\,\mathrm{Wb}} \]

Step 1: Identify the angle.

The angle is measured from the plane, so:

\[ \alpha=30^\circ \]

Step 2: Select the correct equation.

\[ |\Phi_B|=BA\sin\alpha \]

Step 3: Substitute the values.

\[ |\Phi_B|=0.50\times0.018\times\sin30^\circ \]

\[ =0.50\times0.018\times0.5 \]

\[ =0.0045\,\mathrm{Wb} \]

The flux magnitude is \(4.5\times10^{-3}\,\mathrm{Wb}\).

The important step was recognising that the given angle was measured from the plane, so sine was appropriate.

Question 2

Two coils, A and B, are placed in the same uniform magnetic field of \(0.40\,\mathrm{T}\).

Their properties are:

Calculate the flux linkage of each coil and determine which coil has the greater flux linkage.

Assume that the chosen positive normal for each coil points in a direction with a positive component of the magnetic field.

Solution 2

Coil B has the greater flux linkage:

\[ \boxed{\lambda_A=0.16\,\mathrm{Wb\text{-}turn}} \]

\[ \boxed{\lambda_B=0.20\,\mathrm{Wb\text{-}turn}} \]

Step 1: Calculate the flux through one turn of coil A.

The angle is given to the plane, so:

\[ \alpha_A=30^\circ \]

Using:

\[ \Phi_A=BA\sin\alpha_A \]

\[ =0.40\times0.010\times\sin30^\circ \]

\[ =0.0020\,\mathrm{Wb} \]

Step 2: Calculate the flux linkage of coil A.

\[ \lambda_A=N_A\Phi_A \]

\[ =80\times0.0020 \]

\[ \lambda_A=0.16\,\mathrm{Wb\text{-}turn} \]

Step 3: Calculate the flux through one turn of coil B.

This question gives the angle to the normal, so we can use the cosine equation directly:

\[ \Phi_B=BA\cos60^\circ \]

\[ =0.40\times0.020\times0.5 \]

\[ =0.0040\,\mathrm{Wb} \]

Step 4: Calculate the flux linkage of coil B.

\[ \lambda_B=N_B\Phi_B \]

\[ =50\times0.0040 \]

\[ \lambda_B=0.20\,\mathrm{Wb\text{-}turn} \]

Coil B has the greater flux linkage, even though it has fewer turns.

The reason is its larger area. Flux linkage depends on the combined effect of the number of turns, the area, and the orientation of the coil.

The tempting shortcut is to compare the number of turns alone. That ignores the fact that each turn of coil B encloses twice the area of a turn of coil A.

Question 3

A 100-turn flat coil has an area of \(0.015\,\mathrm{m^2}\). It is placed in a uniform magnetic field of \(0.40\,\mathrm{T}\).

A measurement gives the magnitude of its magnetic flux linkage as \(0.36\,\mathrm{Wb\text{-}turn}\).

Calculate the acute angle between the magnetic field and the plane of the coil.

Can the measurement also tell you whether the signed magnetic flux linkage is positive or negative? Explain.

Solution 3

The acute angle between the magnetic field and the coil’s plane is:

\[ \boxed{\alpha=36.9^\circ} \]

The measurement does not tell us the sign of the flux linkage.

Step 1: Start with the flux linkage equation.

\[ |\lambda|=NBA\sin\alpha \]

We need to rearrange this equation to find the angle.

\[ \sin\alpha=\frac{|\lambda|}{NBA} \]

Step 2: Substitute the known values.

\[ \sin\alpha= \frac{0.36}{100\times0.40\times0.015} \]

\[ \sin\alpha=\frac{0.36}{0.60} \]

\[ \sin\alpha=0.60 \]

Step 3: Calculate the angle.

\[ \alpha=\sin^{-1}(0.60) \]

\[ \boxed{\alpha=36.9^\circ} \]

Step 4: Consider the sign.

The measurement provides only the magnitude of the flux linkage.

It tells us how much magnetic flux links the coil, but not which direction the magnetic field points relative to the chosen area normal.

To determine the sign, we would need to know the direction of the chosen area vector and the direction of the magnetic field.

This is an important distinction: a magnitude measurement cannot, by itself, establish a vector’s direction.

Question 4

A 40-turn coil has a constant area of \(0.020\,\mathrm{m^2}\). It is placed in a uniform magnetic field.

The coil is initially positioned so that:

• The magnetic field strength is \(0.30\,\mathrm{T}\).
• The angle between the field and the coil’s plane is \(60^\circ\).

The coil is then rotated to a new position where:

• The magnetic field strength increases to \(0.45\,\mathrm{T}\).
• The angle between the field and the coil’s plane decreases to \(30^\circ\).

The magnetic field direction remains fixed, and the coil does not flip its area normal.

Calculate the initial and final flux linkages. Determine whether the flux linkage increases or decreases, and calculate the percentage change.

Solution 4

The flux linkage decreases by approximately:

\[ \boxed{13.4\%} \]

Despite the increase in magnetic field strength, the change in orientation causes the overall flux linkage to decrease.

Step 1: Calculate the initial flux linkage.

Use:

\[ \lambda=NBA\sin\alpha \]

Initially:

\[ N=40,\quad B_1=0.30\,\mathrm{T} \]

\[ A=0.020\,\mathrm{m^2},\quad \alpha_1=60^\circ \]

Substitute:

\[ \lambda_1= 40\times0.30\times0.020 \times\sin60^\circ \]

\[ \lambda_1=0.24\times0.8660 \]

\[ \boxed{\lambda_1=0.2078\,\mathrm{Wb\text{-}turn} } \]

Step 2: Calculate the final flux linkage.

The new field strength is \(0.45\,\mathrm{T}\), and the new angle to the plane is \(30^\circ\).

\[ \lambda_2=NBA\sin\alpha_2 \]

\[ =40\times0.45\times0.020 \times\sin30^\circ \]

\[ =0.36\times0.5 \]

\[ \boxed{\lambda_2=0.1800\,\mathrm{Wb\text{-}turn}} \]

Step 3: Determine the change in flux linkage.

\[ \Delta\lambda=\lambda_2-\lambda_1 \]

\[ =0.1800-0.2078 \]

\[ \Delta\lambda=-0.0278\,\mathrm{Wb\text{-}turn} \]

The negative sign tells us that the positive flux linkage has decreased.

Step 4: Calculate the percentage decrease.

Use the initial flux linkage as the reference:

\[ \%\text{ decrease}= \frac{\lambda_1-\lambda_2}{\lambda_1} \times100 \]

\[ =\frac{0.2078-0.1800}{0.2078} \times100 \]

\[ \boxed{13.4\%\text{ decrease}} \]

The interesting part is that the magnetic field increased by 50%, but the orientation factor decreased from \(\sin60^\circ\) to \(\sin30^\circ\).

The two effects compete. The reduction in the perpendicular component of the field is large enough to outweigh the increase in field strength.

This is why you should never assume that a stronger magnetic field automatically means greater magnetic flux. Orientation matters just as much.

Question 5

A 50-turn rectangular coil has a constant area of \(0.010\,\mathrm{m^2}\).

Initially:

• The uniform magnetic field has a strength of \(0.40\,\mathrm{T}\).
• The angle between the magnetic field and the coil’s plane is \(30^\circ\).
• The chosen area vector points in a direction that gives positive magnetic flux.

During the next \(0.30\,\mathrm{s}\), two things happen simultaneously:

• The coil rotates through \(180^\circ\) about an axis lying in its own plane.
• The magnetic field strength decreases to \(0.20\,\mathrm{T}\), while its direction remains fixed.

At the end of the rotation, the coil’s plane is geometrically in the same orientation as it was initially. However, its area vector now points in the opposite direction.

Calculate:

a) The signed magnetic flux through one turn initially and finally.

b) The initial and final signed flux linkages.

c) The change in flux linkage.

d) The magnitude of the average induced EMF over the \(0.30\,\mathrm{s}\) interval.

Assume the field is uniform across the coil at all times, and the coil’s area and number of turns remain constant.

Solution 5

The magnitude of the average induced EMF is:

\[ \boxed{|\mathcal E_{\mathrm{avg}}|=0.50\,\mathrm{V}} \]

The crucial detail is that the area vector reverses direction. Although the coil returns to the same geometric plane, its signed magnetic flux changes from positive to negative.

Step 1: Calculate the initial magnetic flux.

Initially:

\[ B_i=0.40\,\mathrm{T} \]

\[ A=0.010\,\mathrm{m^2} \]

\[ \alpha=30^\circ \]

The area vector initially points in the positive direction relative to the magnetic field, so the flux is positive.

\[ \Phi_i=+B_iA\sin30^\circ \]

\[ =+0.40\times0.010\times0.5 \]

\[ \boxed{\Phi_i=+0.0020\,\mathrm{Wb}} \]

Step 2: Calculate the final magnetic flux.

At the end, the magnetic field strength has decreased to \(0.20\,\mathrm{T}\).

However, the coil has also rotated through \(180^\circ\), reversing its area vector.

The flux is therefore negative:

\[ \Phi_f=-B_fA\sin30^\circ \]

\[ =-0.20\times0.010\times0.5 \]

\[ \boxed{\Phi_f=-0.0010\,\mathrm{Wb}} \]

Notice that the angle between the field and the plane is still \(30^\circ\). The negative sign comes from the reversal of the area vector, not from a change in the angle to the plane.

Step 3: Calculate the initial and final flux linkages.

Initially:

\[ \lambda_i=N\Phi_i \]

\[ =50\times0.0020 \]

\[ \boxed{\lambda_i=+0.100\,\mathrm{Wb\text{-}turn}} \]

Finally:

\[ \lambda_f=N\Phi_f \]

\[ =50\times(-0.0010) \]

\[ \boxed{\lambda_f=-0.050\,\mathrm{Wb\text{-}turn}} \]

Step 4: Calculate the change in flux linkage.

Always subtract the initial value from the final value:

\[ \Delta\lambda=\lambda_f-\lambda_i \]

\[ =-0.050-(+0.100) \]

\[ \boxed{\Delta\lambda=-0.150\,\mathrm{Wb\text{-}turn}} \]

The change is \(0.150\,\mathrm{Wb\text{-}turn}\) in magnitude.

Step 5: Calculate the average induced EMF.

Faraday’s law gives the magnitude of the average induced EMF:

\[ |\mathcal E_{\mathrm{avg}}| =\frac{|\Delta\lambda|}{\Delta t} \]

Substitute:

\[ |\mathcal E_{\mathrm{avg}}| =\frac{0.150}{0.30} \]

\[ \boxed{|\mathcal E_{\mathrm{avg}}|=0.50\,\mathrm{V}} \]

The average induced EMF has a magnitude of \(0.50\,\mathrm{V}\).

Why is this question tricky?

A tempting approach is to notice that the magnetic field strength halves and conclude that the flux linkage simply halves.

That would give a change in flux linkage of:

\[ 0.100-0.050=0.050\,\mathrm{Wb\text{-}turn} \]

But this compares magnitudes, not signed values.

The actual final linkage is negative, so the change is:

\[ -0.050-(+0.100)=-0.150 \]

The magnitude of the actual change is three times larger than the tempting answer.

Here’s the deeper idea: the geometric plane of a coil does not uniquely specify its area vector.

A flat rectangle looks geometrically identical after a \(180^\circ\) rotation about an axis in its plane. However, its oriented normal has reversed.

If a question gives you only the angle between the magnetic field and the plane at two different times, you may not have enough information to determine the signed change in flux. You must also know how the area vector is oriented.

That distinction becomes essential when calculating induced EMF using Faraday’s law.

08What to learn next

Once you can calculate the correct magnetic flux and flux linkage at each instant, the next step is understanding how quickly that linkage changes.

That’s what determines the magnitude of the induced EMF.

Work through How to Read a Magnetic-Flux-Time Graph to Calculate EMF to connect the orientation calculations in this article to gradients, changing flux linkage, and Faraday’s law.

If you want more practice with how the enclosed area affects magnetic flux, see How to Compare Magnetic Flux Through Circular and Square Coils.