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Figure 7 shows a search coil positioned on the axis of an electromagnet, with the plane of the search coil perpendicular to the axis - AQA - A-Level Physics - Question 4 - 2021 - Paper 2

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Figure 7 shows a search coil positioned on the axis of an electromagnet, with the plane of the search coil perpendicular to the axis. A magnetic field is produced by... show full transcript

Worked Solution & Example Answer:Figure 7 shows a search coil positioned on the axis of an electromagnet, with the plane of the search coil perpendicular to the axis - AQA - A-Level Physics - Question 4 - 2021 - Paper 2

Step 1

The search coil is placed at $x = 0.070 \, m$.

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Answer

To find the magnetic flux linkage Phi\\Phi through the search coil, we use the formula:

Φ=N×A×B\Phi = N \times A \times B

where:

  • N=200N = 200 (number of turns)
  • A=3.5×10−5 m2A = 3.5 \times 10^{-5} \, m^{2} (cross-sectional area)
  • For x=0.070 mx = 0.070 \, m, refer to Figure 8 for BB, which is found to be 0.005 T0.005 \, T.

Plugging in the values:

Φ=200×3.5×10−5×0.005=4.9×10−4 Wb\Phi = 200 \times 3.5 \times 10^{-5} \times 0.005 = 4.9 \times 10^{-4} \, Wb

Step 2

The search coil is now moved at a constant speed of $0.80 \, m \, s^{-1}$ along the axis so that $x$ is increasing.

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As the coil moves away from the electromagnet, the distance xx increases. The change in distance results in a change in the magnetic flux density BB, resulting in the rate of change of flux linkage, rac{d\Phi}{dt}, which induces an emf according to Faraday's law:

emf=−dΦdt\text{emf} = -\frac{d\Phi}{dt}

Thus, as the search coil continues to move, the flux linkage decreases, inducing an emf that varies depending on the rate of change of flux.

Step 3

The search coil passes through the position where $x = 0.10 \, m$.

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At this position, we need to assess the magnetic flux density BB again from Figure 8. We establish whether the induced emf can be greater than 5 V5 \, V. The emf can be calculated using the relationship from Faraday's law, which states that the maximum induced emf arises when the rate of change of flux linkage is at its peak. If we calculate the value of BB at x=0.10 mx = 0.10 \, m and apply it to the formula, we can determine if emf can exceed 5 V5 \, V based on the induced conditions.

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