## 2 Design and Fabrication of the Current Leads
### Vapor-Cooled Current Lead
The shape ratio values indicate that the cross-sectional area of the brass current lead was approximately nine times that of copper. \[Q_{0}=\frac{h_{L}\,\widetilde{k}}{c_{p0}}\bigg{(}\frac{A}{l}\bigg{)}\ln\bigg{[} \frac{c_{p0}(T_{l}-T_{0})}{h_{L}}+1\bigg{]} \tag{7}\]  
The minimum conductive heat input (\(Q_{0}\)) at the cold end through the vapor-cooled current lead can be calculated through Equation (7) [18]. When copper RRR10, RRR 30, and brass (Cu 60: Zn 40) are used as the constituent materials of the vapor-cooled current lead, the conducted heat leakages per single current lead from room temperature to the temperature of liquid helium are 0.0294 W, 0.0190 W and 0.0549 W, respectively. When the current lead's warm and cold ends are 300 K and 4 K, a conductive heat leakage with respect to the length of the current lead as a function of material is shown in Figure 4. The amount of conductive heat leakage decreases as the length of the current lead increases. When a 1 m length of copper (RRR10) is used as a metal current lead, conductive heat leakage is 0.057 W. When the length of the current lead is 0.98 m, the conductive heat leakage attributed to the two current leads is 0.0588 W for copper (RRR 10), 0.0380 W for copper (RRR 30), and 0.1098 W for brass (Cu 60:Zn 40). The heat generation of the current lead by electrical current should be considered. According to [PERSON]'s law, the resistance and heat generation of current lead are proportional to each other. Therefore, material with relatively low resistance should be selected. As a result of analyses with Equations (5)-(7), it is advantageous in terms of thermal and shape optimization to use copper as a material for vapor-cooled current leads rather than brass. Detailed calculation result values of current lead with respect to the material are summarized in Table 2. Accordingly, brass was not considered in the fabrication process of the current lead. We measured the thermal conductivity of the copper used  
Figure 3: Thermal properties of metals as a function of temperature. in this system. In the previous study [22], the measurement system of thermal contact conductance and thermal conductivity was developed, and the thermal conductivity of copper test sample was obtained by measuring the temperature deviation that occurred when a specific amount of heat was applied to the test sample. The measured results are shown in \[Q_{out}=\left[Ak(T)\frac{dT}{dz}\right]_{z}+\dot{m}_{I}C_{p}(T)(T+\Delta T) \tag{2}\]0 As indicated, the measured values were in close agreement with the RRR 10 value. Therefore, the thermal properties of copper RRR 10 were used for the design of the current lead. There are various sources of heat leakage in the superconducting magnet system, including the instrumental wire conduction, support conduction, liquid helium vessel wall conduction, radiation, residual gas conduction, and heat leakage owing to current lead. Among these, heat leakage by current leads, conductive heat leakage by liquid helium vessel neck, and conductive heat leakage by residual gas account for approximately 90 % of the total heat leakage. When a 45 A electrical current is supplied through the current lead, the total amount of heat leakage of the superconducting magnet is 0.4322 W, and the liquid helium boil-off rate is predicted to be approximately 14.31 L/day. The heat leakage through current lead is 0.114 W accounting for 26% of the total heat leakage, which represents the largest proportion among each heat source. Therefore, the boil-off rate of liquid helium through the current lead must be minimized. \[Q_{out}=\left[Ak(T)\frac{dT}{dz}\right]_{z}+\dot{m}_{I}C_{p}(T)(T+\Delta T) \tag{2}\]1 Heat leakage through the current lead composed of different materials as a function of its length. \[Q_{out}=\left[Ak(T)\frac{dT}{dz}\right]_{z}+\dot{m}_{I}C_{p}(T)(T+\Delta T) \tag{2}\]2 Thermal conductivity of copper with respect to temperature at different copper purity values.