An aeroplane, with its wings spread 10 m, is flying at a speed of \[180\]km/h in a horizontal direction. The total intensity of earth's field at that part is \[2.5 \times {10^{ - 4}}{\text{ Wb/}}{{\text{m}}^2}\], and the angle of dip is \[{60^o}\]. The emf induced between the tips of the plane wings will be:
A. \[88.37\]mV
B. \[62.50\]mV
C. \[54.125\]mV
D. \[108.25\]mV

Hint:Here, in this question we need to determine the induced emf between the tips of the wings of the aeroplane. For this, we will use the formula of induced emf using scalar triple products such as magnetic field, length and velocity.

Formula used:
The emf induced by movement with respect to the magnetic field is,
\[e = \left[ {\vec B\vec V\vec L} \right]\]
\[e = BVL \times \sin \theta \]
where,
\[e\] is the induced emf, \[B\] is the magnetic field, \[V\] is the velocity of an object, \[L\] is the length and \[\theta \] is the angle between \[V\] and \[B\].

Complete step by step solution:
We know that the induced emf between the tips of the plane wings is,
\[e = \left[ {\vec B\vec V\vec L} \right]\]
Let us use the formula for the scalar triple product of vectors.
We know that \[\left[ {\vec a\vec b\vec c} \right] = ab \times \sin \theta \]
Accordingly, the induced emf is given by, \[e = B \times V \times L \times \sin \theta \].
We know that \[V = 180\]km/h, \[\theta = {60^o}\], \[B = 2.5 \times {10^{ - 4}}{\text{ Wb/}}{{\text{m}}^2}\], and \[L = 10\]m.

Here, we need to change the velocity in metre per seconds as-
\[v = \dfrac{{180 \times 1000}}{{60 \times 60}} \\
v= 180 \times \dfrac{5}{{18}}\,{\text{m/sec}} \\ \]
By substituting these values in the above formula, we get
\[e = 2.5 \times {10^{ - 4}} \times \left( {180 \times \dfrac{5}{{18}}} \right) \times 10 \times \sin \left( {{{60}^o}} \right)\]
Thus, after simplification, we get
\[e = 108.25 \times {10^{ - 3}}{\text{ }}V\]
\[\therefore e = 108.25{\text{ m}}V\]
Therefore, the emf induced between the tips of the wings of an aeroplane is \[108.25{\text{ }}mV\].
Hence, the option (D) is correct.

Note:Here is one more concept of solving this question. We can also write the formula of emf as \[e = {B_v}vl\] where \[{B_v}\] is the vertical component of the magnetic field.

Image: vector representation of scalar product

So,
\[{B_v} = B\sin 60 \\
{B_v} = \dfrac{{\sqrt 3 B}}{2} \\ \]
Now, substituting the known parameters in the formula \[e = {B_v}vl\] as-
\[e = {B_v}vl \\
\Rightarrow e= \dfrac{{\sqrt 3 }}{2} \times 2.5 \times {10^{ - 4}} \times \left( {180 \times \dfrac{5}{{18}}} \right) \times 10 \\
\therefore e= 108.25\,mV \]