A Hall probe consists of copper strip of n=8.5 times 10^28 mathrm(~m)^-3 , Shich is 2.0 mathrm(~cm) of Cross-seetinal dimensims 2.0 mathrm(~cm) by 0.10 mathrm(~cm) . Determine tho magnetro field when I=50 mathrm(~A) and V_(H)=4.0 mathrm(MV) .

To determine the magnetic field (\(B\)) in the Hall probe, we can use the Hall effect equation:<br /><br />\[ V_H = \frac{B \cdot l \cdot v}{n \cdot e} \]<br /><br />Where:<br />- \(V_H\) is the Hall voltage,<br />- \(B\) is the magnetic field,<br />- \(l\) is the length of the Hall probe,<br />- \(v\) is the drift velocity of the charge carriers,<br />- \(n\) is the number of charge carriers per unit volume,<br />- \(e\) is the charge of the electron.<br /><br />Given:<br />- \(V_H = 4.0 \, \text{MV} = 4.0 \times 10^6 \, \text{V}\)<br />- \(I = 50 \, \text{A}\)<br />- \(n = 8.5 \times 10^{28} \, \text{m}^{-3}\)<br />- \(e = 1.6 \times 10^{-19} \, \text{C}\)<br /><br />First, we need to find the drift velocity (\(v\)) using the formula:<br /><br />\[ v = \frac{I}{n \cdot A \cdot e} \]<br /><br />Where:<br />- \(A\) is the cross-sectional area of the Hall probe.<br /><br />Given the dimensions of the Hall probe:<br />- Width (\(w\)) = \(2.0 \, \text{cm} = 0.02 \, \text{m}\)<br />- Height (\(h\)) = \(0.10 \, \text{cm} = 0.001 \, \text{m}\)<br /><br />The cross-sectional area (\(A\)) is:<br /><br />\[ A = w \cdot h = 0.02 \, \text{m} \cdot 0.001 \, \text{m} = 2 \times 10^{-5} \, \text{m}^2 \]<br /><br />Now, calculate the drift velocity (\(v\)):<br /><br />\[ v = \frac{I}{n \cdot A \cdot e} = \frac{50 \, \text{A}}{8.5 \times 10^{28} \, \text{m}^{-3} \cdot 2 \times 10^{-5} \, \text{m}^2 \cdot 1.6 \times 10^{-19} \, \text{C}} \]<br /><br />\[ v = \frac{50}{8.5 \times 10^{28} \cdot 2 \times 10^{-5} \cdot 1.6 \times 10^{-19}} \]<br /><br />\[ v = \frac{50}{2.72 \times 10^4} \]<br /><br />\[ v = 1.84 \times 10^{-3} \, \text{m/s} \]<br /><br />Now, we can find the magnetic field (\(B\)) using the Hall effect equation:<br /><br />\[ B = \frac{V_H \cdot n \cdot e}{l \cdot v} \]<br /><br />Given the length of the Hall probe (\(l\)) is \(2.0 \, \text{cm} = 0.02 \, \text{m}\), we can substitute the values into the equation:<br /><br />\[ B = \frac{4.0 \times 10^6 \, \text{V} \cdot 8.5 \times 10^{28} \, \text{m}^{-3} \cdot 1.6 \times 10^{-19} \, \text{C}}{0.02 \, \text{m} \cdot 1.84 \times 10^{-3} \, \text{m/s}} \]<br /><br />\[ B = \frac{4.0 \times 10^6 \cdot 8.5 \cdot 1.6}{0.02 \cdot 1.84} \]<br /><br />\[ B = \frac{5.76 \times 10^{10}}{0.0368} \]<br /><br />\[ B = 1.56 \times 10^9 \, \text{T} \]<br /><br />Therefore, the magnetic field when \(I = 50 \, \text{A}\) and \(V_H = 4.0 \, \text{MV}\) is \(1.56 \times 10^9 \, \text{T}\).