Calculating Hall Voltage with Given Values | Electromagnetism Problem"

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The discussion focuses on calculating the Hall voltage using given parameters: magnetic field (B = 2.5 T), current (I = 15 A), charge of an electron (q = 1.6 x 10^-19 C), and electron density (n = 8.47 x 10^22 electrons/cm^3) for a slab of specific dimensions. The formula used is V_H = (IBd) / (nqA), where v_d is derived from the current and density. The calculated Hall voltage is approximately 1.3 microvolts, leading to uncertainty about the multiple-choice answers provided. The participant suspects their calculation may contain an error since their result does not match any of the options. The conclusion indicates a need for re-evaluation of the calculations to align with the expected answers.
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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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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