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Chapter 26: Problem 53

A rectangular copper strip measures \(1.0 \mathrm{mm}\) in the direction of a uniform 2.4 -T magnetic field. When the strip carries a \(6.8-\) A current perpendicular to the field, a 1.2 - \(\mu\) V Hall potential develops across the strip. Find the number density of free electrons in the copper.

Short Answer

Expert verified
The number density of free electrons in the copper is \(8.8 * 10^{28} m^{-3}\).

Step by step solution

01

Identify the given values

Let's review what is given: magnetic field strength (B) = 2.4 T, current (I) = 6.8 A, Hall potential (V_H) = 1.2 μV and thickness (d) = 1.0 mm. Since the Hall voltage needs to be consistent with the other units, it should to be in volts. Hence, V_H = 1.2 μV = \(1.2 * 10^{-6}\) V. Also, the thickness (d) should be in meters, i.e., d = 1.0 mm = 0.001 meter.
02

Use the formula for the Hall Coefficient

The Hall coefficient (R_H) can be found using the relation \(R_H = \frac{V_H * d}{B * I}\). By using the given values and making the substitutions to get: \(R_H = \frac{(1.2 * 10^{-6} V) * (0.001 m)}{(2.4 T) * (6.8 A)} = 7.14 * 10^{-11} m^3/C\).
03

Find the number density of free electrons

We can use the relation between the number density of free electrons (n) and Hall coefficient \(R_H\), i.e, \(n = -1 / (e * R_H)\). Here, the factor -1 is due to the negative charge of the electron and e is the charge of an electron = \(1.6 * 10^{-19}\) C. Substituting the values gives \(n = -1 / ((1.6 * 10^{-19} C) * (7.14 * 10^{-11} m^3/C)) = 8.8 * 10^{28} m^{-3}\). The negative sign can be ignored as it was just taking into account the direction of the electrons.

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