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Using information in Example $20.6,$ what would the Hall voltage be if a 2.00 - field is applied across a 10 -gauge copper wire $(2.588 \mathrm{mm}$ in diameter) carrying a $20.0-\mathrm{A}$ current?

The Hall voltage is $1.47 \times 10^{-6} \mathrm{V}$

Physics 102 Electricity and Magnetism

Chapter 22

Magnetism

Magnetic Field and Magnetic Forces

Sources of Magnetic field

Electromagnetic Induction

Inductance

Cornell University

Simon Fraser University

Hope College

University of Sheffield

Lectures

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The electric force is a ph…

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Figure 26.27 gives the ele…

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A 14-gauge copper wire has…

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Calculate the current dens…

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we know that the formula to find the drift velocity feats of D would be equaling I, which would be the electric current divided by the number of electrons per cubic meter and a times A the cross sectional area of the wire multiplied by Q. The charge of an electron. We know that a the cross sectional area is equal in Thai R squared. This would be equally high multiplied by we have 2.588 millimeters or we can say 0.2 588 meters divided by two. So the diameter divided by two quantity squared. And so the cross sectional area of the wire is found to be 5.26 times 10 to the negative six square meters. Now we can use this velocity in order to find the velocity formula, and the current, we know is 20 points zero amperes divided by 8.34 times 10 to the 28th electrons per cubic meter multiplied by the charge of one electron, 1.60 times 10 to the negative 19th cool arms multiplied by 5.26 times 10 to the negative six meters squared and we find that the drift velocity is equaling 2.849 times, 10 to the negative forthe meters per second. And so we can say that then calculating the whole voltage would be simply the product between the magnitude of the magnetic field. The length of the water of the wire multiply rather out, not like the the width of the conductor. My apology multiplied by the drift velocity. And so here we have a ah magnetic field of 2.0 Tesla's The width of the conductor is 2.588 times 10 to the negative third meters. So essentially the diameter of 2.588 millimeters multiplied bye and we have 2.849 times 10 to the negative fourth meters per second for the drips velocity. And so we find not the whole voltage produced would be 1.47 times 10 to the negative sixth volts. This would be our final answer. That is the end of the solution. Thank you for watching

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