Question

# a.) I recently was on an airplane, and I was thinking about the fact that a...

a.) I recently was on an airplane, and I was thinking about the fact that a Boeing 747 is basically a giant flying conductor surrounded by the Earth's magnetic field. The plane's motion through the Earth's magnetic field should induce an EMF between the wing tips of the plane, which are about 60 meters apart. Suppose that the plane's speed parallel to the ground is 800 km/hr and that the Earth's magnetic field has a strength of about 0.3 Gauss (a Gauss is 10−410−4 T). In the northern hemisphere, the magnetic field tends to tilt at an angle of about 60 degrees relative to the Earth's surface. Just how big is the induced EMF across the plane?

b.) I've got a uniform magnetic field of strength 100 mT. How quickly (in terms of frequency) must I spin a coil with resistance 10 mΩΩ and a radius of 2 cm in order to produce a 100 mA RMS-averaged current? Assume that the coil starts with its area fully exposed to the magnetic field.

Solution) L = 60 m

V = 800 km/h

1 km/h = (1000)/(3600) m/s

1 km/h = (5/18) m/s

V = 800(5/18) = 222.2 m/s

B = 0.3 Gauss

1 Gauss = 10^(-4) T

B = 0.3×10^(-4) T

Theeta = 60°

Emf , E = ?

E = d(phi)/dt

phi = B.A = B.(L.X)cos(theeta)

E = B.Lsin(theeta)(d(X)/dt) = BLVsin(theeta)

E = 0.3×10^(-4)×60×222.2×sin(60)

E = 0.346 V

(b) B = 100 mT = 100×10^(-3) T

R = 10 m ohms = 10×10^(-3) ohms

r = 2 cm = 2×10^(-2) m

I(rms) = 100 mA = 100×10^(-3) A

w = ?

I(peak) = 1.414×I(rms)

I(peak) = 1.414×100×10^(-3)

I(peak) = 141.4×10^(-3) A

I(peak) = E/R

E = - d(phi)/dt

phi = B.A = B.Acos(wt)

d(phi)/dt = B.A[d(cos(wt))/dt]

E = BAwsin(wt)

For Emax , sin(wt) = 1

E(peak) = BAw

A = (pi)(r^2)

A = (pi)(2×10^(-2))^2 = 12.56×10^(-4) m^2

I(peak) = (BAw)/(R)

I(peak) = (100×10^(-3)×12.56×10^(-4)×w)/(10×10^(-3))

141.4×10^(-3) = 0.01256(w)

w = (141.4×10^(-3))/(0.01256)

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