Question

A hypothetical metal is known to have an electrical resistivity of 3.3×10−8(Ω⋅m). A current of 25 A is passed through a specimen of this metal 15 mm thick. When a magnetic field of 0.95 tesla is simultaneously imposed in a direction perpendicular to that of the current, a Hall voltage of −2.4×10−7V is measured. Compute the following:
(a) the electron mobility for this metal
(b) the number of free electrons per cubic meter

Answer 1

Answer:

a) The electron mobility for this metal is 0.00455 m²V/s

b) There are 4.12 * 10^28 free electrons per cubic meter

Explanation:

Step 1: Data given

electrical resistivity of 3.3 * 10^−8(Ω⋅m)

A current of 25 A is passed through a specimen of this metal 15 mm thick.

A magnetic field of 0.95 tesla.

A Hall voltage of −2.4×10−7V is measured

Step 2: Calculate the conductivity

σ= 1/ ρ

 ⇒ with ρ = electrical resistivity of 3.3*10^−8Ω*m

σ = 1/3.3*10^−8

σ = 30303030.3 = 3*10^7/ Ω*m

step 3: Calculate the electron mobility

µe = Vdσ / IxBz

⇒ with µe = the electron mobility in m²V/s

⇒ V = voltage of  2.4*10^−7V

⇒ d = 0.015 meter

⇒ σ = the conductivity = 3*10^7/ Ω*m

⇒ Ix = the current = 25 A

⇒ Bz =  magnetic field = 0.95 tesla

µe = (2.4 * 10^-7 * 0.015 * 3*10^7) / (25*0.95)

µe = 0.00455 m²V¨/s

The electron mobility is 0.00455 m²V¨/s

Step 4: Calculate the number of free electrons per cubic meter

n = σ/e*µe

⇒ e= the charge of an electron = 1.6 * 10^-19 Coulombs

⇒ σ = The conductivity = 3*10^7/ Ω*m

⇒ µe = The electron mobility = 0.00455 m²V¨/s

n = 3*10^7/ Ω*m /(1.6 * 10^-19 * 0.00455)

n = 4.12 * 10^28 / m³

There are 4.12 * 10^28 free electrons per cubic meter