---
title: "V3Ga critical temperature and fields"
subject: "Electrical Engineering Material"
chapter: 5
key: "semester_3/electrical_engineering_material:v3ga-critical-temperature"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/v3ga-critical-temperature
site: BCT Engineering
---

# V3Ga critical temperature and fields

Electrical Engineering Material · Chapter 5: Superconductivity

- Asked in: 2073 Shrawan, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/v3ga-critical-temperature
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-5

## Question

For a specimen of V₃Ga, the critical fields are 0.176T and 0.528T for 14K and 13K respectively. Calculate the critical temperature. Also calculate critical fields at 0K and 4.2K.

## Short answer

Using H_c(T)=H_0[1−(T/T_c)^2] and the data at 14 K (0.176 T) and 13 K (0.528 T), solving the ratio yields T_c=14.47 K, H_c(0)=2.73 T, and H_c(4.2 K)=2.50 T.

## Answer

The critical field varies with temperature as

$$
H_c(T) = H_0\left[1 - \frac{T^2}{T_c^2}\right]
$$

**Given:** $H_{c1} = 0.176$ T at $T_1 = 14$ K, $H_{c2} = 0.528$ T at $T_2 = 13$ K.

#### Critical temperature
Divide the two equations:

$$
\begin{aligned} \frac{0.176}{0.528} &= \frac{T_c^2 - 14^2}{T_c^2 - 13^2} \\ \frac{1}{3} &= \frac{T_c^2 - 196}{T_c^2 - 169} \\ T_c^2 - 169 &= 3T_c^2 - 588 \\ 2T_c^2 &= 419 \\ T_c^2 &= 209.5 \\ T_c &= 14.47\ \text{K} \end{aligned}
$$

#### Critical field at 0 K
$$
\begin{aligned} H_0 &= \frac{H_c(14)}{1 - \frac{196}{209.5}} = \frac{0.176 \times 209.5}{13.5} \\ &= 2.731\ \text{T} \end{aligned}
$$

Check with the 13 K value: $2.731 \times (1 - 169/209.5) = 2.731 \times 0.1933 = 0.528$ T, which matches.

#### Critical field at 4.2 K
$$
\begin{aligned} H_c(4.2) &= 2.731\left[1 - \frac{4.2^2}{209.5}\right] \\ &= 2.731\,(1 - 0.0842) \\ &= 2.501\ \text{T} \end{aligned}
$$

**Answer:** $T_c = 14.47$ K, $H_c(0) = 2.73$ T, $H_c(4.2\ \text{K}) = 2.50$ T.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is Tc found from two (T, Hc) data points?
   A. Divide the two equations Hc = H0[1 - T^2/Tc^2] to eliminate H0 and solve for Tc^2
   B. Add the two fields and divide by two
   C. Take the average of 13 K and 14 K
   D. Use Tc = T1 + T2 directly
   Answer: A. The ratio 0.176/0.528 = (Tc^2 - 196)/(Tc^2 - 169) removes H0. (Key point: Method for Tc)

2. What is the critical temperature of V3Ga here?
   A. 13.5 K
   B. 14.47 K
   C. 15.2 K
   D. 209.5 K
   Answer: B. 1/3 = (Tc^2 - 196)/(Tc^2 - 169) gives Tc^2 = 209.5, so Tc = 14.47 K. (Key point: Value of Tc)

3. What is H0 (the critical field at 0 K)?
   A. About 0.176 T
   B. About 0.528 T
   C. About 2.73 T
   D. About 14.5 T
   Answer: C. H0 = 0.176 x 209.5 / (209.5 - 196) = 2.731 T. (Key point: Hc at 0 K)

4. What is Hc at 4.2 K?
   A. About 2.73 T
   B. About 1.30 T
   C. About 0.53 T
   D. About 2.50 T
   Answer: D. Hc = 2.731 x (1 - 17.64/209.5) = 2.731 x 0.9158 = 2.50 T. (Key point: Hc at 4.2 K)

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