91Ó°ÊÓ

.There is the maximum current in a straight, $3.0-mm$-diameter superconducting niobium wire.

A current-carrying wire produces a magnetic field and the Biot-savart law enables us to calculate the magnitude and direction of this magnetic field. at any point where the magnetic field due to the segment$∆\stackrel{→}{s}$of current- carrying wire is given by equation $\left(29.7\right)$

$B=\frac{{\mathrm{μ}}_{\mathrm{ο}}I}{2\mathrm{π}r}$

Our target is to find the current I, so we solve equation $\left(1\right)$for I to be in form

$I=\frac{2\mathrm{π}Br}{{\mathrm{μ}}_{\mathrm{ο}}}$

Where $r$is the radius of the wire and it is calculated by

$r=\frac{d}{2}=\frac{3mm}{2}=1.5mm$

Now, we plug the values for $B,r$and ${\mathrm{μ}}_{\mathrm{ο}}$into equation $\left(2\right)$to get the maximum current

$$\begin{gathered} I=\frac{2\mathrm{π}Br}{{\mathrm{μ}}_{\mathrm{ο}}} \\ =\frac{2\mathrm{π}\left(0.1T\right)\left(1.5×{10}^{-3}m\right)}{4\mathrm{π}×{10}^{-7}T·m/A} \\ =750A \end{gathered}$$