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Figure 2 shows a modern version of the apparatus used by Hertz to investigate the properties of electromagnetic waves - AQA - A-Level Physics - Question 2 - 2018 - Paper 7

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Figure 2 shows a modern version of the apparatus used by Hertz to investigate the properties of electromagnetic waves. Electromagnetic waves are continuously emitted... show full transcript

Worked Solution & Example Answer:Figure 2 shows a modern version of the apparatus used by Hertz to investigate the properties of electromagnetic waves - AQA - A-Level Physics - Question 2 - 2018 - Paper 7

Step 1

Sketch a graph on Figure 4 to show how the amplitude detected by the dipole receiver varies with angle of rotation as the receiver is turned through 360°

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Answer

The graph should start at its maximum amplitude when the receiver is aligned with the transmitter (0 degrees). As the receiver is rotated to 90 degrees, the amplitude decreases to zero because the dipole receiver is perpendicular to the electromagnetic wave. From 90 degrees to 180 degrees, the amplitude again increases to a maximum, following the same pattern, before dropping back to zero at 270 degrees. Finally, the graph returns to maximum amplitude at 360 degrees, completing the cycle.

This will create a sinusoidal wave-like pattern representing the amplitude versus angle of rotation, with maximum amplitudes at 0°, 180°, and 360° and minimums at 90° and 270°.

Step 2

Explain, using a suitable calculation, why this equation leads to the conclusion that light is an electromagnetic wave.

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Answer

Using the equation derived by Maxwell:

c=1μ0ϵ0c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}
We can substitute the known values for μ0 \mu_0 (permeability of free space) and ϵ0 \epsilon_0 (permittivity of free space):

μ0=4π×10−7 H/m\mu_0 = 4\pi \times 10^{-7} \, \text{H/m}
ϵ0=8.85×10−12 F/m\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}
Substituting these values into the equation, we find:

c=1(4π×10−7)(8.85×10−12)=3.0×108 m/sc = \frac{1}{\sqrt{(4\pi \times 10^{-7})(8.85 \times 10^{-12})}} = 3.0 \times 10^8 \, \text{m/s}

This value matches the measured speed of light, supporting the conclusion that light is indeed an electromagnetic wave.

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