Question

Question: The magnetic field at the surface of a long wire radius R and carrying a current I is $\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{\pi }\mathbf{R}}$ . How large acurrent could a 0.1 mm diameter niobium wire carry without exceeding its 0.2 T critical field?

Short answer

Answer

Current in the wire is 50A.

Step-by-step solution

Given data

Diameter of niobium wire is, 0.1mm .

Maximum magnetic field is,0.2T .

 Step 2: Formula of Magnetic Field produced by a Wire

given by, $\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{\pi }\mathbf{R}}$ .

Here I represents current, R represents radius of wire and${\mathbf{\mu }}_{\mathbf{0}}$ represents permeability of free space.

Calculate the Current from the Expression of Magnetic Field produced by a Wire

The radius of the wire is, $R=\frac{D}{2}$.

Substitute $0.1\text{mm}\to D$ and convert it into m .

$$\begin{gathered} R=\frac{D}{2} \\ =\frac{0.1\text{mm}}{2} \\ =0.05\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right) \\ =0.05\times {10}^{-3}\text{m} \end{gathered}$$

The magnetic field produced by the current is given by, $B=\frac{{\mu }_{0}I}{2\pi R}$.

Rearrange the above equation for l .

$B=\frac{{\mu }_{0}I}{2\pi R}$

Substitute $4\pi \times {10}^{-7}\text{T}·\text{m}/\text{A}\to {\mu }_0,0.2\text{T}\to \text{B}$, and $0.05\times {10}^{-3}\text{m}\to \text{R}$ in the above equation.

$$\begin{gathered} B=\frac{{\mu }_{0}I}{2\pi R} \\ I=\frac{2\pi RB}{{\mu }_0} \\ =\frac{2\pi \left(0.05\times {10}^{-3} \text{m}\right)(0.2 \text{T})}{4\pi \times {10}^{-7} \text{Tm/A}} \\ =50 \text{A} \end{gathered}$$

Current in the wire is 50 A .