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).
Show Hint
Start from \(V_H = R_H I B / t\) and solve for \(B\). Keep every quantity in base SI units before you divide.
Step 1: Compute the Hall resistance first.
Instead of plugging everything into one formula, define the Hall resistance as the ratio of the measured Hall voltage to the current:
\[ R_{Hall} = \frac{V_H}{I} = \frac{0.9 \times 10^{-3}}{6 \times 10^{-3}} = 0.15\ \Omega \]
Step 2: Relate the Hall resistance to the magnetic field.
From $V_H = R_H I B / t$, dividing both sides by $I$ gives $R_{Hall} = R_H B / t$, so
\[ B = \frac{R_{Hall}\, t}{R_H} \]
Step 3: Substitute the numbers.
With $R_{Hall} = 0.15\ \Omega$, $t = 0.7 \times 10^{-3}$ m and $R_H = 4 \times 10^{-4}\ \text{m}^3/\text{C}$:
\[ B = \frac{0.15 \times 0.7 \times 10^{-3}}{4 \times 10^{-4}} = \frac{1.05 \times 10^{-4}}{4 \times 10^{-4}} = 0.2625 \text{ T} \]
Final Answer:
Rounded off to two decimal places, the incident magnetic field is $0.26$ T.
\[ \boxed{B = 0.26 \text{ T}} \]
