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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.

Respuesta :

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

Answer:

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

Ver imagen AFOKE88

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