Question

A fighter plane of length 20 m, wingspan (distance from the tip of one wing to the tip of the other wing) of 15 m and height 5 m is flying towards the east over Delhi. Its speed is $240m{s^{ - 1}}$. The earth's magnetic field over Delhi is $5 \times {10^{ - 5}}T$ with the declination angle ${0^0}$ and dip of θ such that sinθ=$\dfrac{2}{3}$. 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
A. ${V_B} = 40mV;{V_W} = 135mV$with left side of pilot at higher voltage
B. ${V_B} = 45mV;{V_W} = 120mV$ with right side of pilot at higher voltage
C. ${V_B} = 40mV;{V_W} = 135mV$with right side of pilot at higher voltage
D. ${V_B} = 45mV;{V_W} = 120mV$with left side of pilot at higher voltage.

Answer 1

Hint: In electromagnetic induction when a conductor is moving with a certain velocity in the presence of a magnetic field then the voltage will be generated across the conductor and this is called the induced emf. This happens due to a change in the flux which is associated with that conductor. This problem is related to the induced emf.
Formula used:
$\eqalign{
  & \phi = B.A \cr
  & e = BVL\sin \theta \cr} $

Complete step-by-step solution:
The flux which is associated with the wings of the plane will be changing as it is moving with velocity V in the magnetic field B and the length of the wings be L. It is well known that the earth has a magnetic field associated with it. It has both horizontal and vertical components of the magnetic field at every location and the resultant of the horizontal component and vertical component at a particular place gives us the resultant magnetic field at that location. The dip is the angle between the resultant magnetic field vector and the horizontal component vector. The horizontal component of the magnetic field always points in the north direction.
$\phi = B.A$
Where $\phi $ is the magnetic flux associated and A is the area that is changing. The above expression is a dot product of the magnetic field and area.
We have a term called the motional emf which is induced by the motion of a body in a constant magnetic field. The emf generated in this case is motional emf and its formula generally is given by $e = BVL$
Here e is the induced motional emf and it is to be noted that the B and V and L should be mutually perpendicular to each other.
In the question it is given that plane is traveling in the east direction and since wings will be perpendicular to the plane traveling direction it means wings length is along the north direction and we know earth’s horizontal magnetic field also will be in the north direction and hence we should consider earth’s vertical component of a magnetic field which will be $B\sin \theta $ where $\theta $ is the dip angle.
So we have emf generated across wings will be which will be
${V_W} = BVL\sin \theta = 5 \times {10^{ - 5}}T \times 240 \times 15 \times \dfrac{2}{3} = 120mV$
Here the length of the wings should be taken and by using trigonometric identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ we had found the value of $\cos \theta $
${\sin ^2}\theta + {\cos ^2}\theta = 1$
$\eqalign{
  & \Rightarrow {\left[ {\dfrac{2}{3}} \right]^2} + {\cos ^2}\theta = 1 \cr
  & \Rightarrow {\cos ^2}\theta = 1 - \dfrac{4}{9} \cr
  & \Rightarrow {\cos ^2}\theta = \dfrac{5}{9} \cr
  & \Rightarrow \cos \theta = \dfrac{{\sqrt 5 }}{3} \cr} $
Now due to the horizontal component of magnetic field emf will be generated across the lower and upper side of the plane. Horizontal component of earth’s magnetic field will be $B\cos \theta $. Emf generated across the height will be
${V_B} = BVL\cos \theta = 5 \times {10^{ - 5}}T \times 240 \times 5 \times \dfrac{{\sqrt 5 }}{3} = 45mV$
Now force across the wings will be
$F = q\left( {\mathop V\limits^ \to \times \mathop B\limits^ \to } \right)$
For wings velocity of plane is in positive x-direction and the vertical magnetic field is in negative y-direction. So if we do cross product the positive charge will accumulate at the left side of the wings. So the left side of the pilot will be at higher voltage.
Hence option D will be the correct answer.

Note: If the emf generated is due to the change in a magnetic field then the emf induced may be varying or constant but emf generated due to motion is not varying and can be considered as a DC source. Because as the magnetic field is varying the rate of change of the magnetic field also will be varying. If that rate is constant then that emf also can be considered as constant emf.