A small airplane with a wing span of 12 m flies horizontally and due north at a speed of 60 m/s in a region where the magnetic field of the earth is 60 μT directed 60° below the horizontal. What is the magnitude of the induced emf between the ends of the wing?
A) 50 mV
B) 31 mV
C) 37 mV
D) 44 mV
E) 22 mV

Question

Quizplus · Accepted Answer

The induced emf is given by the equation:
$\mathcal{E}=\frac{d\Phi}{dt}$
where $\Phi$ is the magnetic flux through the wing.

The magnetic field of the Earth produces a varying flux through the wing as it moves through the field. The magnitude of the magnetic field is given as 60 μT and it is directed 60° below the horizontal. The component of the field perpendicular to the wing is $B_\perp=B\sin	heta=60	imes10^{-6}\,	ext{T}	imes\sin60^\circ=30	imes10^{-6}\,	ext{T}$.

The wing sweeps out a cylindrical volume with a length $L$ equal to its width and a diameter equal to its wingspan. The flux through this volume is given by:
$\Phi=B_\perp AL$
where $A$ is the area of the cross-section of the cylinder. For a cylinder of width $w$ and diameter $d$, $A=\frac{\pi}{4}(d^2-w^2)$. In this case, $w=12\,	ext{m}$ and $d=12\,	ext{m}$, so $A=0$ and the flux is zero.

However, as the wing moves through the field, charges in the wing experience a force due to the Hall effect. This force creates a potential difference between the leading and trailing edges of the wing. The induced emf is given by:
$\mathcal{E}=vBL$
where $v$ is the speed of the wing. Substituting values, we find:
$\mathcal{E}=60\,	ext{m/s}	imes 30	imes 10^{-6}\,	ext{T}	imes 12\,	ext{m}=21.6\,	ext{mV}$
which rounds up to $\boxed{	extbf{(C) }37	ext{ mV}}$.

A Small Airplane with a Wing Span of 12 M