Solution 1: Aircraft in Earth’s Magnetic Field
1. Analysis of Motional EMF:
When an aircraft flies horizontally, its wings cut the vertical component of the Earth’s magnetic field ($B_v$). This creates a motional electromotive force (EMF) across the wingtips.
The magnitude of this induced voltage is given by:
$$ \varepsilon = B_v \cdot L \cdot v $$Where:
- $B_v$ is the vertical component of Earth’s magnetic field.
- $L$ is the wingspan (length of the conductor).
- $v$ is the speed of the aircraft.
Thus, a voltage will be induced across the wings. One wingtip will be at a higher potential than the other.
Figure 1: The wing acts as a battery, but the connected circuit moves with it through the uniform field.
2. Will the bulb glow?
For the electric bulb to glow, a steady current must flow through its filament. A current requires a net electromotive force (EMF) around the closed loop formed by the wings and the connecting wires.
Consider the closed loop formed by the wing and the external wire connecting the bulb:
- The entire system (wing + wires + bulb) is moving with the same velocity $v$ through a uniform magnetic field.
- The magnetic flux $\Phi_B$ passing through this loop is constant because the magnetic field $B$ is uniform and the area of the loop is not changing.
According to Faraday’s Law of Induction:
$$ \varepsilon_{\text{net}} = -\frac{d\Phi_B}{dt} $$Since $\Phi_B$ is constant, $\frac{d\Phi_B}{dt} = 0$. Therefore, the net EMF in the loop is zero, and no current flows.
Conclusion: While a potential difference exists across the wingtips (due to motional EMF), there is no net EMF to drive current through the bulb circuit because the connecting wires act as identical “batteries” opposing the wing’s EMF.