Question

A Boeing 747 with a 65m65-\mathrm{m} wingspan is cruising northward at 250m/s250 \mathrm{m} / \mathrm{s} toward Alaska. The B\vec{B} field at this location is 5.0×105T5.0 \times 10^{-5} \mathrm{T} and points 6060^{\circ} below its direction of travel. Determine the potential difference between the tips of its wings.

Solution

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We know that the induced emf is equal to the negative rate change of magnetic flux and expression is given by:

ε=ΔϕΔt=BLvcosθ\varepsilon=\bigg |\dfrac{\Delta \phi}{\Delta t}\bigg | =BLv \cos \theta

Now plug in the required value in the above equation and solve

ε=(5105 T)65 m(250 ms)cos30°=0.704 V\begin{align*} \varepsilon&=(5 \cdot 10^{-5} \ \text{T})65 \ \text{m} (250 \ \frac{\text{m}}{\text{s}}) \cdot \cos 30\text{\textdegree}\\ &=0.704 \ \text{V} \end{align*}

ε=0.704 V\boxed{\color{#c34632}\varepsilon=0.704 \ \text{V}}

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