A thin ribbon of copper 6 mm high and 0.5 mm deep, carrying a current of 20 ampe

Question

A thin ribbon of copper 6 mm high and 0.5 mm deep, carrying a current of 20 amperes is placed in a magnetic field of 1 tesla. For copper, q = -e, and n = 8.4 1028 free electrons per m3 (one per atom).

Once the circuit has reached a steady state, which of the following are true?

There is a non-zero electric field in the direction of conventional current.

There is an induced electric field in the opposite direction as the magnetic field.

There is a net upward magnetic force on the charge carriers.

This situation will never reach a steady state.

There is an upward electric force on the charge carriers that exactly cancels the downward magnetic force.

The conventional current I=20 amperes is given by

i = |q|vn(hd)uEi = AuE    I = nAuEI = |q|n(hd)v

Calculate the Hall voltage VHall for this ribbon of copper
V
This is a very small voltage, but it can be measured with a sensitive voltmeter. In semiconductors the density of charge carriers n may be quite small compared to a metal, in which case the Hall-effect voltage can be quite large, since it is inversely proportional to n.

Explanation / Answer

V(H) = -iB / dnE
i = current=20A
B = magnetic field=1Tesla
d = depth of plate=0.5mm=0.5*10^-3m
n = free electrons per m^3=8.4*10^28
e = charge=1.6*10^-19
V(H) = -iB / dnE=-(20*1) / ((0.5*10^-3)*(8.4*10^28)*(1.6*10^-19)) = -2.98*10^-6V

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