Question

Calculate the potential difference induced between the tips of the wings of a Boeing \(747-400\) with a wingspan of \(64.67 \mathrm{~m}\) when it is in level flight at a speed of \(913 \mathrm{~km} / \mathrm{h}\). Assume that the magnitude of the downward component of the Earth's magnetic field is \(B=5.00 \cdot 10^{-5} \mathrm{~T}\). a) \(0.820 \mathrm{~V}\) b) \(2.95 \mathrm{~V}\) c) \(10.4 \mathrm{~V}\) d) \(30.1 \mathrm{~V}\) e) \(225 \mathrm{~V}\)

Short answer

Answer: The potential difference between the tips of the wings of the Boeing 747-400 is a) \(0.820 \mathrm{~V}\).

Step-by-step solution

Convert speed to meters per second

The plane's speed is given in km/h, but we need to convert it into meters per second (m/s) to work with the rest of the values provided. To do this, use the conversion factor: \(1 \mathrm{~km/h} = \frac{1000}{3600} \mathrm{~m/s}\). Multiply the plane's speed by this factor: \(913 \mathrm{~km/h} \cdot \frac{1000}{3600} \mathrm{~m/s} = 253.61 \mathrm{~m/s}\).

Calculate the EMF in the wings

Now, we will use the formula for the EMF to find the induced potential difference between the wingtips. The formula is: \(EMF = BLv\). Plug in the provided values: \(EMF = (5.00 \cdot 10^{-5} \mathrm{~T})(64.67 \mathrm{~m})(253.61 \mathrm{~m/s})\).

Solve for the potential difference

Multiply the values in the equation to find the induced potential difference: \(EMF = (5.00 \cdot 10^{-5} \mathrm{~T})(64.67 \mathrm{~m})(253.61 \mathrm{~m/s}) = 0.82 \mathrm{~V}\). The potential difference (or induced potential difference) between the tips of the wings of a Boeing 747-400 flying at the given speed and altitude is: a) \(0.820 \mathrm{~V}\).