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# An aeroplane having a distance of 50 m between the edges of its wings is flying horizontally with a speed of 360 km/hr. If the vertical components of earth’s magnetic field is $2 \times {10^{ - 4}}Wb/{m^2}$, then the induced emf between the edges of its wings will be A. 1VB. 3VC. 0.2VD. 12V

Last updated date: 16th Jun 2024
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Hint:In order to solve this question, the concept of motional emf must be understood. Emotional emf is defined as the emf induced in the metal conductor when it is in movement by cutting across a magnetic field in the direction normal to that of the magnetic field.

The Faraday’s law of electromagnetism states that:
Whenever there is change in the magnetic flux linked with the closed loop conductor, there is an emf induced at its end.
$Emf,E = - \dfrac{{d\phi }}{{dt}}$
where $\dfrac{{d\phi }}{{dt}}$= rate of change of magnetic flux in unit area.
Consider an aeroplane with the length of the wings 50 m flying with a speed, $v = 360kmph = 360 \times \dfrac{5}{{18}} = 100m{s^{ - 1}}$
An emf is induced when the earth’s magnetic field induced flux linked with the aeroplane wings changes due to movement of the aeroplane.
This emf induced is called motional emf.
Motional emf, $E = - \dfrac{{d\phi }}{{dt}}$
(here, the negative sign is only indicative of the direction and not the true value)
However, the above equation gives us the emf induced per unit length of the wings. Since the length of the wings is given as, $l = 50m$
$E = l\dfrac{{d\phi }}{{dt}} = l\dfrac{d}{{dt}}\left( {B \times A} \right)$
$E = Bl\dfrac{{dA}}{{dt}}$

Here, the rate of change of area swept over time is equal to the velocity of the aeroplane. Thus –
$\dfrac{{dA}}{{dt}} = v$
$E = Bl\dfrac{{dA}}{{dt}}$
Substituting, we get –
$E = Blv$
Given the magnetic field component, $B = 2 \times {10^{ - 4}}Wb/{m^2}$
$E = Blv$
Substituting,
$E = 2 \times {10^{ - 4}} \times 50 \times 100 = 1V$
Hence, the motional emf induced, E = 1V

Hence, the correct option is Option A.

Note:At this point, some of you may wonder how it is possible to multiply two vector quantities like magnetic field B and velocity V without considering any directions. Here is how:
In actuality, it is the cross-product of the two vectors.
$\overrightarrow B \times \overrightarrow v = \overrightarrow {\left| B \right|} \overrightarrow {\left| v \right|} \sin \theta$
Here, in the question, it is mentioned as the vertical component. So, I have considered
$\left| {\overrightarrow B } \right| = B\sin \theta$
$\left| {\overrightarrow B } \right| = B\sin 90 = B$
Hence, the value of the field is directly chosen that way.