JEE Main 2016 · 10 Apr · Shift 1 · Physics Q16

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PhysicsSingle correct+4 / -1Question 16 of 90

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 ms $^{-1}$ . The earth's magnetic field over Delhi is $5\times10^{-5}$ T with the declination angle ~0° and dip of θ such that $\sin\theta=\dfrac23$ . 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 :

V_B=45

mV;

V_W=120

mV with right side of pilot at higher voltage

V_B=45

mV;

V_W=120

mV with left side of pilot at higher voltage

V_B=40

mV;

V_W=135

mV with right side of pilot at high voltage

V_B=40

mV;

V_W=135

mV with left side of pilot at higher voltage

Answer and worked solution

Official answer: B

$\cos\theta=\frac{\sqrt5}3$ .

Between the lower and upper sides (height 5 m) the horizontal component acts: $V_B=B\cos\theta\,v\,h=5\times10^{-5}\times0.745\times240\times5\approx45$ mV.
Between the wing tips (15 m) the vertical component acts: $V_W=B\sin\theta\,v\,l=5\times10^{-5}\times\frac23\times240\times15=120$ mV.

Over Delhi the vertical component points down. Flying east, $\vec v\times\vec B$ points north, which is the pilot's left: the left wing is at the higher potential.

Topics:Motional EMF Earth's magnetism

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