Use the following figure for the next problem. -A metal strip 1.5 cm wide and 0.20 cm thick carries a current of 15 A in a uniform magnetic field of 1.5 T. The Hall emf is 5.3 $\mu$V. The electronic charge is 1.6 * 10^-19 C. The number density of the charge carriers in the strip is approximately

Question

Quizplus · Accepted Answer

The Hall voltage (V_H) is given by V_H = (IB)/(nq), where I is the current, B is the magnetic field, n is the charge carrier density, and q is the charge of an electron. Rearranging for n gives n = (IB)/(V_H * q). Substituting the given values: I = 15 A, B = 1.5 T, V_H = 5.3 x 10^-6 V, and q = 1.6 x 10^-19 C, we find n ≈ 1.3 x 10^28 electrons/m^3.

Use the Following Figure for the Next Problem μ\muμ V the Electronic Charge Is 1

Use the Following Figure for the Next Problem $\mu$ V the Electronic Charge Is 1