Calculating Hall Voltage with Given Values | Electromagnetism Problem"

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SUMMARY

The Hall voltage for the given parameters is calculated using the formula V_H = (IBd) / (nqA). With a magnetic field (B) of 2.5 T, current (I) of 15 A, charge (q) of 1.6 x 10^-19 C, and electron density (n) of 8.47 x 10^22 electrons/cm^3, the calculated Hall voltage is approximately 1.3 µV. The dimensions of the slab are 0.20 cm thick and 1.5 cm wide, leading to a cross-sectional area (A) of 0.0003 m². The calculated value does not match any of the multiple-choice answers provided, indicating a potential error in the calculations or assumptions made.

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Homework Statement



What is the Hall voltage? B = 2.5 T, I = 15 A, q = 1.6*10^-19 C, n = 8.47*10^22 electrons/cm^3. B is going through a square slab 0.20 cm x 1.5 cm (B going through 0.20 cm thick part... not sure which one though counts as the width)

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Homework Equations


V_H = v_d * B * w where w is width
v_d = I/(nqA)
Thus,
V = (IBd) / (nqA)

The Attempt at a Solution


V = (15 A * 2.5 T * 0.015 m) / ((8.47*10^22 electrons/cm^3)*(1*10^6 cm^3 / m^3)*(1.6*10^-19 C)*(0.015 m * 0.0020 m) = 1.3*10^-6V

It is multiple choice and the answers are 2.6 V, 1.9 mV, 3.6 micro V, 4.2 mV, or none of them. I am leaning towards none of the above but it usually is pretty rare for one of these questions so I suspect there's an error in my calculations...
 
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I get about same answer, 1.4 micro V
 

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