If the number of electrons per unit volume in the copper is 8.46 × 10²⁸ m⁻³ and the current in the strip is 19.0 A, calculate the potential difference across the width of the strip when conditions have been allowed to equilibrate.

The first part of the question is missing and it says;

Electrons are made to flow through the copper strip of as shown in the attached image . The strip's cross section is 1.00 mm high and 30.5 mm wide, and the strip is placed in a magnetic field B = 4.00T

Hall potential difference = 5.6 x 10^(-6) V

Explanation:

We are given that;

Thickness of copper strip; L = 1mm = 1 x 10^(-3)m

Width of copper strip; W = 30.5mm

Magnetic field; B = 4T

Current in the strip = 19A

Number density; n = 8.46 × 10²⁸ m⁻³

Formula for number density of charge is given by;

n = (B•I)/(V•L•e)

Where;

B is uniform Magnetic field

I is current

V is potential difference

L is thickness of strip

e is electron charge = 1.6 x 10^(-19)C

Thus, let's make V the subject;

V = (B•I)/(n•L•e)

V = (4 x 19)/(8.46 × 10²⁸•1 x 10^(-3)•1.6 x 10^(-19))

V = 76/(13.536 x 10^(6)) = 5.6 x 10^(-6) V

snell's law

refractve index = sin incidence / sin refraction