GATE 2026 IN Q.63: solution and answer

GATE 2026 · INQ.63 · 2 marks · NAT

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 x 10^-4 m^3/C, the value of the incident magnetic field is ______ T. (rounded off to two decimal places)

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Answer and solution

ANSWER

The incident magnetic field is $B \approx 0.26 T$ .

Official answer: 0.25 to 0.27Source: Official answer key, GATE 2026 IN (IIT Guwahati)

For a slab of thickness $t$ measured along the magnetic field, carrying current $I$ , the Hall voltage is $V_{H} = \frac{R _{H} I B}{t}$ .
Solve for the field: $B = \frac{V _{H} t}{R _{H} I}$ .
Convert to SI units: $V_{H} = 0.9 \times 1 0^{- 3} V$ , $t = 0.7 \times 1 0^{- 3} m$ , $I = 6 \times 1 0^{- 3} A$ , $R_{H} = 4 \times 1 0^{- 4} m^{3} / C$ .
Substitute: $B = \frac{0.9 \times 1 0 ^{- 3} \times 0.7 \times 1 0 ^{- 3}}{4 \times 1 0 ^{- 4} \times 6 \times 1 0 ^{- 3}} = \frac{6.3 \times 1 0 ^{- 7}}{2.4 \times 1 0 ^{- 6}} = 0.2625 T$
Rounded to two decimal places, $B \approx 0.26 T$ .

KEY IDEA

Use the Hall relation $V_{H} = R_{H} I B / t$ , where $t$ is the sample dimension along the field. Rearrange to $B = V_{H} t / (R_{H} I)$ and convert every quantity to SI units (mV to V, mm to m, mA to A) before substituting.

WATCH OUT FOR

Leaving mm, mV or mA unconverted, which shifts the answer by powers of ten. Another slip is writing $V_{H} = R_{H} I B / w$ with the width instead of the thickness along the field. The thickness is the only length given here.

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GATE 2026 IN Q.63: solution and answer

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