A fighter plane of length 20, m, wing span (d | vedclass

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Question

(A) The earth's magnetic field $B = 5 	imes 10^{-5}\,T$. The vertical component is $B_V = B \sin 	heta = 5 	imes 10^{-5} 	imes (2/3) \approx 3.33

Vedclass · Accepted Answer

$V_B = 40\, mV$; $V_W = 135\,mV$ with left side of pilot at higher voltage

Vedclass · Answer

(A) The earth's magnetic field $B = 5 	imes 10^{-5}\,T$. The vertical component is $B_V = B \sin 	heta = 5 	imes 10^{-5} 	imes (2/3) \approx 3.33 	imes 10^{-5}\,T$.The horizontal component is $B_H = B \cos 	heta = B \sqrt{1 - \sin^2 	heta} = 5 	imes 10^{-5} 	imes \sqrt{1 - 4/9} = 5 	imes 10^{-5} 	imes \sqrt{5}/3 \approx 3.73 	imes 10^{-5}\,T$.For the voltage $V_B$ between the lower and upper side (height $h = 5\,m$), the plane moves horizontally, so the vertical component $B_V$ cuts the height: $V_B = B_V \cdot v \cdot h = (3.33 	imes 10^{-5}) 	imes 240 	imes 5 = 0.04\,V = 40\,mV$.For the voltage $V_W$ between the wing tips (span $w = 15\,m$), the plane moves east, so the horizontal component $B_H$ (which is north-south) cuts the wings: $V_W = B_H \cdot v \cdot w = (3.73 	imes 10^{-5}) 	imes 240 	imes 15 \approx 0.134\,V = 134\,mV \approx 135\,mV$.Using the right-hand rule for motional $EMF$ $(\vec{v} 	imes \vec{B})$, with velocity towards east and magnetic field towards north, the force on positive charges is towards the left of the pilot.

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