A fighter plane of length 20 m, wing span (distance from tip of one wing to the tip of the other wing) of 15 m and height 5 m is flying towards east over Delhi. Its speed is 240 m s$$^{-1}$$. The earth's magnetic field over Delhi is $$5 \times 10^{-5}$$ T with the declination angle ~0° and dip of $$\theta$$ such that $$\sin \theta = \frac{2}{3}$$. If the voltage developed is $$V_B$$ between the lower and upper side of the plane and $$V_W$$ between the tips of the wings then $$V_B$$ and $$V_W$$ are close to:

We treat the aeroplane as a set of straight metallic conductors moving with uniform velocity in the earth’s magnetic field. Whenever a conductor of length $$L$$ moves with velocity $$\vec v$$ through a magnetic field $$\vec B$$, an emf

$$\varepsilon \;=\; (\vec v \times \vec B)\cdot\vec L$$

is produced, where $$\vec L$$ is a vector from one end of the conductor to the other. The magnitude is therefore

$$\vert\varepsilon\vert = v\,B_\perp\,L = vBL\sin\phi,$$

$$\phi$$ being the angle between $$\vec v$$ and $$\vec B$$.

The data given are

$$\begin{aligned} B &= 5\times10^{-5}\,\text{T},\\ \sin\theta &= \frac23 \;(\text{dip}),\\ v &= 240\ \text{m s}^{-1},\\ \text{height} &= 5\ \text{m},\\ \text{wing span} &= 15\ \text{m}. \end{aligned}$$

Because the declination is $$\approx 0^\circ$$, the horizontal component of $$\vec B$$ is due north. We choose axes

$$\hat i :$$ East $$,\qquad \hat j :$$ North $$,\qquad \hat k :$$ Up $$.$$

With dip $$\theta$$, the magnetic field is

$$\vec B = B\cos\theta\,\hat j - B\sin\theta\,\hat k.$$

Using $$\sin\theta=\dfrac23,\; \cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\dfrac49}= \sqrt{\dfrac59}= \dfrac{\sqrt5}{3},$$ we have

$$\begin{aligned} B_H &= B\cos\theta = 5\times10^{-5}\times\frac{\sqrt5}{3}\ \text{T},\\ B_V &= B\sin\theta = 5\times10^{-5}\times\frac23\ \text{T}. \end{aligned}$$

Numerically

$$\begin{aligned} B_H &= 5\times10^{-5}\times0.745355\;=\;3.72678\times10^{-5}\ \text{T},\\[2mm] B_V &= 5\times10^{-5}\times0.666667\;=\;3.33333\times10^{-5}\ \text{T}. \end{aligned}$$

The velocity of the aircraft is due east, $$\vec v = v\,\hat i$$. Hence

$$\vec v\times\vec B = v\$$, $$\hat i \times \bigl(B_H\hat j - B_V\hat k\bigr) = vB_H(\hat i\times\hat j) - vB_V(\hat i\times\hat k) = vB_H\$$, $$\hat k - vB_V\$$, $$\hat j.$$

Thus

$$\vec v\times\vec B = vB_H\,\hat k \;-\; vB_V\,\hat j.$$

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Emf between lower and upper surfaces (height 5 m)

The length vector from the lower surface to the upper surface is $$\vec L_B = 5\,\hat k\ \text{m}.$$ Therefore

$$\varepsilon_B = (\vec v\times\vec B)\cdot\vec L_B = \bigl(vB_H\,\hat k - vB_V\,\hat j\bigr)\cdot (5\,\hat k) = vB_H\times5.$$

Substituting numbers,

$$\begin{aligned} \varepsilon_B &= 240\,(3.72678\times10^{-5})\times5 \\[2mm] &= 240\times5\times3.72678\times10^{-5}\\[2mm] &= 1200\times3.72678\times10^{-5}\\[2mm] &= 4.47214\times10^{-2}\ \text{V}\\[2mm] &\approx 0.045\ \text{V}\;=\;45\ \text{mV}. \end{aligned}$$

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Emf between the wing tips (span 15 m)

The wings run north-south, the left wing tip being to the north of the pilot. We take the length vector from the right tip (south) to the left tip (north):

$$\vec L_W = 15\,\hat j\ \text{m}.$$

Hence

$$\varepsilon_W = (\vec v\times\vec B)\cdot\vec L_W = \bigl(vB_H\,\hat k - vB_V\,\hat j\bigr)\cdot(15\,\hat j) = -\,vB_V\times15.$$

Its magnitude is

$$\vert\varepsilon_W\vert = vB_V\times15 = 240\,(3.33333\times10^{-5})\times15.$$

Step by step:

$$\begin{aligned} 240\times15 &= 3600,\\ 3600\times3.33333\times10^{-5} &= 1.20000\times10^{-1}\ \text{V}\\ &= 0.12\ \text{V}\\ &= 120\ \text{mV}. \end{aligned}$$

The negative sign in $$\varepsilon_W$$ shows that the potential decreases from north to south, i.e. the left (north) wing tip is at a higher potential than the right (south) wing tip.

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We have therefore obtained

$$V_B \approx 45\$$ mV $$,\qquad V_W \approx 120\$$ mV $$,$$

with the left side of the pilot (north wing tip) at the higher voltage.

Hence, the correct answer is Option D.