A neutral metal rod of length 52 cm falls toward the Earth. The rod is horizonta

Question

A neutral metal rod of length 52 cm falls toward the Earth. The rod is horizontal and oriented east-west. At a moment when the rod's speed is 3 m/s, approximately how many excess electrons are at the negative end of the rod. (Take the Earth's magnetic field to be 2*10^-5 T).

A neutral metal rod of length 52 cm falls toward the Earth. The rod is horizontal and oriented east-west. At a moment when the rod's speed is 3 m/s, approximately how many excess electrons are at the negative end of the rod. (Take the Earth's magnetic field to be 2*10^-5 T).

Explanation / Answer

Emf induced in rod e = BVL

e = 2*10^-5 *3 *0.52

e = 3.12*10^-5 volt

now if electric field is E then

qE = kq^2 / L^2

qe/L = Kq^2/L^2

q = Le/K = (0.52x3.12*10^-5)/9x10^9

q = 1.8x10^-15 C

number of electron n = q/1.6x10^-19

n = 1.125x10^4 electrons

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