A square coil of wire is rotating at a constant angular speed about a horizontal axis. Figure 4 shows the coil at one instant when the normal to the plane of the coil is at 30° to a magnetic field. The area of the coil is 5.0 × 10⁻⁴ m² and the flux density of the uniform magnetic field is 2.5 × 10² T. The maximum flux linkage of the coil during its rotation is 1.5 × 10³ Wb turns. Calculate the number of turns in the coil. Calculate the flux linkage of the coil at the instant shown in Figure 4. The coil forms part of an electrical generator. Figure 5 shows the emf generated by the coil. Calculate the peak value of the emf generated. Sketch on Figure 6 the variation with time of flux linkage for the same time interval as Figure 5.

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A square coil of wire is rotating at a constant angular speed about a horizontal axis - AQA - A-Level Physics - Question 5 - 2019 - Paper 2

Question 5

A square coil of wire is rotating at a constant angular speed about a horizontal axis. Figure 4 shows the coil at one instant when the normal to the plane of the coi... show full transcript

Worked Solution & Example Answer:A square coil of wire is rotating at a constant angular speed about a horizontal axis - AQA - A-Level Physics - Question 5 - 2019 - Paper 2

Step 1

Calculate the number of turns in the coil.

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Answer

To find the number of turns, use the formula for magnetic flux linkage:

$\Phi = N A B \cos(\theta)$

Where:

$\,N$ is the number of turns,
$\,A$ is the area of the coil ( $5.0 \times 10^{-4} \ m²$ ),
$\,B$ is the magnetic flux density ( $2.5 \times 10^{2} \, T$ ),
$\,\theta$ is the angle (30°).

First, calculate the maximum flux linkage: $\Phi_{max} = A imes B$ $\Phi_{max} = (5.0 \times 10^{-4} \ m²) \times (2.5 \times 10^{2} \, T) = 1.25 \times 10^{-1} \ Wb$

Using the maximum flux linkage: $1.5 \times 10^{3} \ Wb = N \times 1.25 \times 10^{-1} \ Wb$

This gives: $N = \frac{1.5 \times 10^{3} \ Wb}{1.25 \times 10^{-1} \ Wb} = 120 \ turns$ .

Step 2

Calculate the flux linkage of the coil at the instant shown in Figure 4.

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Answer

Using the previously mentioned formula for flux linkage:

$\Phi = N A B \cos(30^{\circ})$

Substituting the known values:

$\,N = 120 \ turns$ ,
$\,A = 5.0 \times 10^{-4} \ m²$ ,
$\,B = 2.5 \times 10^{2} \, T$ ,
$\,\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$ .

This results in: $\Phi = 120 \times (5.0 \times 10^{-4}) \times (2.5 \times 10^{2}) \times \frac{\sqrt{3}}{2} = 1.3 \times 10^{3} \ Wb \ turns.$

Step 3

Calculate the peak value of the emf generated.

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Answer

The peak emf (E) generated can be found using Faraday's law: $E = N A B \omega \sin(\theta)$

Where:

$\,\omega$ is the angular velocity (in radians per second). Assuming an angular speed of 1 Hz (for this calculation): $\omega = 2\pi \times 1 \, rad/s$

Substituting known values: $E = 120 \times (5.0 \times 10^{-4} \ m²) \times (2.5 \times 10^{2} \, T) \times (2\pi) = 0.15 \ V.$

Step 4

Sketch on Figure 6 the variation with time of flux linkage for the same time interval as Figure 5.

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Answer

The graph of flux linkage will mirror the emf graph shown in Figure 5, exhibiting a sinusoidal variation. The flux linkage will oscillate between positive and negative peaks, similarly shaped as the emf curve. A solid or dashed line may be used to outline the curve, depicting a sinusoidal wave pattern within the time frame represented.

A square coil of wire is rotating at a constant angular speed about a horizontal axis - AQA - A-Level Physics - Question 5 - 2019 - Paper 2

Worked Solution & Example Answer:A square coil of wire is rotating at a constant angular speed about a horizontal axis - AQA - A-Level Physics - Question 5 - 2019 - Paper 2

Calculate the number of turns in the coil.

Calculate the flux linkage of the coil at the instant shown in Figure 4.

Calculate the peak value of the emf generated.

Sketch on Figure 6 the variation with time of flux linkage for the same time interval as Figure 5.

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