Question

In Problem \(35,\) the width of the strip is \(3.5 \mathrm{cm},\) the magnetic field is \(0.43 \mathrm{T}\), the Hall voltage is measured to be $7.2 \mu \mathrm{V},\( the thickness of the strip is \)0.24 \mathrm{mm},$ and the current in the wire is 54 A. What is the density of carriers (number of carriers per unit volume) in the strip?

Short answer

Answer: The density of carriers in the strip is approximately 2.49 * 10^28 m^-3.

Step-by-step solution

Write down the given information and formula

Write the given information: Width of the strip (w) = 3.5 cm = 0.035 m (convert it to meters) Magnetic field (B) = 0.43 T Hall voltage (V_H) = 7.2 μV = 7.2 * 10^-6 V (convert it to volts) Thickness of the strip (t) = 0.24 mm = 0.00024 m (convert it to meters) Current in the wire (I) = 54 A The Hall effect formula is: V_H = I * B * t / n * e * w Where: V_H = Hall voltage I = Current in the wire B = Magnetic field t = Thickness of the strip w = Width of the strip n = Density of carriers (number of carriers per unit volume, which we need to find) e = Charge of an electron = 1.6 * 10^-19 C (coulombs)

Rearrange the formula for carrier density n

Rearrange the Hall effect formula to solve for n: n = I * B * t / (V_H * e * w)

Plug in the given values and solve for n

Plug in the given values into the rearranged formula: n = (54 A * 0.43 T * 0.00024 m) / (7.2 * 10^-6 V * 1.6 * 10^-19 C * 0.035 m) Now, perform the calculation and find the value of n: n ≈ 2.49 * 10^28 m^-3

Write down the final answer

The density of carriers (number of carriers per unit volume) in the strip is: n ≈ 2.49 * 10^28 m^-3