Physics Asked on November 8, 2021
If the wing of an air plane is cuboidal in shape, and that the Earth’s magnetic field $bf{B}$ is uniform in the neighbouring volume of the plane, does that mean the induced emf over the entire ‘block’ of wing is zero due to
$$Phi=oint_S mathbf{B cdot mathbf{n}}dS=0 tag{1}$$
and hence $E=-frac{d Phi}{dt}=0$?
Therefore only asymmertical wings (which are a common feature found in modern aeroplanes) can give rise to an induced emf due to $bf{B}$?
(a) Your phrase "the induced emf over the entire block of wing doesn't make much sense to me. You need to specify a closed path, along which you calculate the emf. This path need not be through the metal of the wing, but you may wish it to be.
(b) If the closed path moves with the plane, and the plane moves in a straight line through a uniform magnetic field, then $Phi$ in your equation will be a constant (not generally zero), but $mathscr E = frac {dPhi}{dt}=0$ so the emf will still be zero.
(c) This is the case whether or not the wing is symmetrical, as it applies to all closed loops moving with the plane.
(d) Homework and exam questions sometimes ask students to calculate the emf between the wingtips of an aircraft in flight. This emf is zero unless the wingspan of the aircraft in included in a loop not all of which is moving with the aircraft. One crazy arrangement would be to have the wingtips brushing against metal fences anchored to the ground! You could connect a voltmeter between the fences.
Answered by Philip Wood on November 8, 2021
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