Question

An aluminium wire with a square cross-section 4 mm on a side is carrying a current of 4.5 amps. Aluminium has a resistivity of 2.69 × 10−6 ohm-cm and an atomic volume of 10 cc/mol. Assuming aluminium has three free electrons per atom, calculate the (a) Hall coefficient, (b) electric field, (c) power dissipated (heat generated) per cm length, and (d) the Hall voltage across the wire in a transverse magnetic field of 1 tesla.

Answer 1

a) We can find the Hall coefficient from the formula:

$$R_H=\dfrac{1}{\rho}=\dfrac{1}{2.69\cdot10^{-8}\ \Omega\cdot m}=37.17\cdot10^6\ \dfrac{m^3}{As}.$$

b) We can find the electric field from the formula:

$$E=\dfrac{IB}{\rho l^2}=\dfrac{4.5\ A\cdot 1\ T}{2.69\cdot10^{-8}\ \Omega\cdot m\cdot(4.0\cdot10^{-3}\ m)^2}=1.04\cdot10^{13}\ \dfrac{V}{m}.$$

c)-d) Let's firs find the Hall voltage:

$$V_H=\dfrac{IB}{\rho l}=\dfrac{4.5\ A\cdot 1\ T}{2.69\cdot10^{-8}\ \Omega\cdot m\cdot 4.0\cdot10^{-3}\ m}=4.18\cdot10^{10}\ V.$$

Then, we can find the power dissipated:

$$P=IV_H=4.5\ A\cdot 4.18\cdot10^{10}\ V=18.81\cdot10^{10}\ J.$$

Answer:

a) $R_H=37.17\cdot10^6\ \dfrac{m^3}{As}.$

b) $E=1.04\cdot10^{13}\ \dfrac{V}{m}.$

c)-d) $P=18.81\cdot10^{10}\ J, V_H=4.18\cdot10^{10}\ V.$