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3 years ago

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The cross section of a copper strip is 1.2 mmthick and 20 mm wide. There is a 25-A current through this cross section, with the

charge carriers traveling down the length of the strip. The strip is placed in a uniform magnetic field that has a magnitude of 2.5 T and is directed perpendicular to both the length and the width of the strip. The number density of free electrons in copperis 8.47 ×1019mm−3. a. Calculate the speed of the electrons in the strip. b. Calculate the potential difference across the strip width.

1 answer:

Naily [24]3 years ago

5 0

To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

$V_d = \frac{I}{nAq}$

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

$I = 25 A$

$A= 1.2*20 *10^{-6} m^2$

$q= 1.6*10^{-19}C$

$N = 8.47*10^{19} mm^{-3}$

$V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}$

$V_d = 7.68*10^{-5}m/s$

The hall voltage is given by

$V=\frac{IB}{ned}$

Where

B= Magnetic field

n = number of free electrons

d = distance

e = charge of electron

Then using the formula and replacing,

$V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}$

$V = 3.84*10^{-6}V$

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