A thin ribbon of copper 5 mm high and 0.2 mm deep, carrying a current of 23 ampe

Question

A thin ribbon of copper 5 mm high and 0.2 mm deep, carrying a current of 23 amperes is placed in a magnetic field of 3 tesla. For copper, q = -e, and n = 8.4 ? 1028 free electrons per m3 (one per atom).

A thin ribbon of copper 5 mm high and 0.2 mm deep, carrying a current of 23 amperes is placed in a magnetic field of 3 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 an upward electric force on the charge carriers that exactly cancels the downward magnetic force There is an induced electric field in the opposite direction as the magnetic fielod ic force O This situation will never reach a steady state a steady state There is a net upward magnetic force on the charge carriers There is a non-zero electric field in the direction of conventional current The conventional current I-23 amperes is given by 1 = nAuE Calculate the Hall voltage tall for this ribbon of copper to n. ery This is a very sma 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 of charge etal, in it is inversel

Explanation / Answer

V(H) = -iB / dnE
i = current
B = magnetic field
d = depth of plate
n = free electrons per m^3
e = charge

V(H) = -(23*3) / (0.2E-3 * 8.4E28 * 1.6E-19) = -8.55E-6 V

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