Question

A current carrying semiconductor of thickness \(0.7\) mm is placed in a transverse magnetic field. The measured Hall voltage is \(0.9\) mV and the current is \(6\) mA. If the Hall coefficient is \(4 \times 10^{-4}\) \(\text{m}^3/\text{C}\), the value of the incident magnetic field is T (rounded off to two decimal places).

Hint:

Start from \(V_H = R_H I B / t\) and solve for \(B\). Keep every quantity in base SI units before you divide.

Answer 1

Correct Answer: 0.26

Step 1: Recall the Hall effect formula.
For a current-carrying slab of thickness \(t\) sitting in a transverse magnetic field \(B\), the Hall voltage that appears across the slab is
\[ V_H = \frac{R_H\, I\, B}{t} \]
where \(R_H\) is the Hall coefficient of the material and \(I\) is the current through the slab. This follows from the charge carriers being pushed sideways by the magnetic (Lorentz) force until the electric field they build up (the Hall field) balances that force.

Step 2: Rearrange the formula for the magnetic field.
\[ B = \frac{V_H\, t}{R_H\, I} \]

Step 3: Substitute the given values.
Here \(V_H = 0.9\) mV \(= 0.9 \times 10^{-3}\) V, \(t = 0.7\) mm \(= 0.7 \times 10^{-3}\) m, \(I = 6\) mA \(= 6 \times 10^{-3}\) A, and \(R_H = 4 \times 10^{-4}\) \(\text{m}^3/\text{C}\).
\[ B = \frac{(0.9 \times 10^{-3})(0.7 \times 10^{-3})}{(4 \times 10^{-4})(6 \times 10^{-3})} \]
The numerator:
\[ 0.9 \times 10^{-3} \times 0.7 \times 10^{-3} = 0.63 \times 10^{-6} \]
The denominator:
\[ 4 \times 10^{-4} \times 6 \times 10^{-3} = 24 \times 10^{-7} = 2.4 \times 10^{-6} \]
So
\[ B = \frac{0.63 \times 10^{-6}}{2.4 \times 10^{-6}} = 0.2625 \text{ T} \]

Final Answer:
Rounded off to two decimal places, the magnetic field is \(0.26\) T. \[ \boxed{0.26 \text{ T}} \]