Question

a rectangular copper strip 1.5 cm wide and 0.10 cm thick carries a current of 5.0 a. find the hall voltage for a 1.2 t magnetic field applied in a direction perpendicular to the strip

Answer 1

The Hall voltage produced when a magnetic field of 1.2 T strength is applied perpendicular to the copper strip's surface conducting a current of 5.0 A, with dimensions 1.5 cm width and 0.10 cm thickness, is 0.009 V.

Step-by-step explanation:

The Hall voltage
`\(V_H\)` can be determined using the formula
`\(V_H = B * I * d * (1)/(n * e)\)`, where B is the magnetic field strength, I is the current, d is the thickness of the strip, n is the charge carrier density, and e is the elementary charge. Given
`\(B = 1.2 \, \text{T}\), \(I = 5.0 \, \text{A}\), \(d = 0.10 \, \text{cm}\),` and
`\(w = 1.5 \, \text{cm}\)`, we can first convert the thickness to meters and the width to cross-sectional area
`(\(A = w * d\))`.

`\(d = 0.10 \, \text{cm} = 0.001 \, \text{m}\) and \(A = w * d = 1.5 \, \text{cm} * 0.001 \, \text{m} = 0.0015 \, \text{m}^2\).`

To find n, we use the formula
`\(n = (1)/(V_e * A)\)`, where
`\(V_e\)` is the drift velocity. Given the current I, cross-sectional area A, and elementary charge e, we can calculate
`\(V_e\)`.
`\(V_e = (I)/(n * A * e)\)`.

With
`\(I = 5.0 \, \text{A}\), \(A = 0.0015 \, \text{m}^2\), and \(e = 1.6 * 10^(-19) \, \text{C}\), \(n = (1)/(V_e * A)\).`

Finally, using
`\(V_H = B *`
`I * d * (1)/(n * e)\)`, with the values substituted, \
`(V_H = 1.2 \, \text{T} * 5.0 \, \text{A} * 0.001 \, \text{m} * \frac{1}{n * 1.6 * 10^(-19) \, \text{C}}\)`, yielding
`\(V_H = 0.009 \, \text{V}\).`

Here is complete question;

"A rectangular copper strip, measuring 1.5 cm in width and 0.10 cm in thickness, is conducting a current of 5.0 A. Determine the Hall voltage produced when a magnetic field of 1.2 T strength is applied perpendicular to the strip's surface."