An airplane with a wingspan of 18.0 m is flying due north, in the northern hemis

Question

An airplane with a wingspan of 18.0 m is flying due north, in the northern hemisphere, at a speed of 61.6 m/s over a region where the Earth's magnetic field is 2.40 mu T and the dip angle is 30 degree. (a) What potential difference is developed between the airplane's wingtips? (b) Looking up at the plane as it is approaching you, which wingtip is at higher potential? (c) How would the answers to parts (a) and (b) change if the plane were flying due east? (d) Can this emf be used to power a lightbulb in the passenger compartment?

Explanation / Answer

(a) According to Faraday's law,

Emf = -d/dt(Magnetic flux linked)

and magnetic flux = B.A = BAcos(30)

Area of the wings in circular motion is given by,

A= 3.14*9*9 = 254.34 m^2

Time is given by, t = displacement/velocity = 18/61.6 = 0.29 sec

so, potential difference =-2.40*10^-6*254.34*cos(30) / 0.29 = -1.822 mV

(b) Since the potential is decreasing with time, the back wingtip is at higher potential than that of front wingtip.

(c) If the plane is flying towards east,

The dip angle will become, cos(180+30) = cos(210)

So, potential difference = -2.40*10^-6*254.34*cos(210)/0.29 = 1.822 mV

Now, since the potential difference is positive, the front wingtip is at higher potential tha the back wingtip.

(d) We know, Magnetic field is given by, B=4*3.14*10^-7*I/2*3.14*18

so, I = 2.40*10^-6*18/ 2*10^-7 = 216 A

So, To light a bulb of Power = 1W

Potential required = 1/216 = 4.62 mV, which is greater than the emf.

So, this emf can't be used to power a light bulb.

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