Question

What currents are needed to generate the magnetic field strengths of Table 29.1 at a point$1.0\mathrm{cm}$from a long, straight wire?

Short answer

The current required $2.5\mathrm{A},250\mathrm{A},(5\mathrm{to}50)\times {10}^4\mathrm{A},5\times {10}^5\mathrm{A}$respectively.

Step-by-step solution

Step 1; Given information

We have given,

Different strength of the magnetic fields at a distance from the wire is $1\mathrm{cm}.$

We have to find the current required for produce the magnetic field.

Simplify

Since the magnetic field due to long wire is given by,

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}$

Where

$$\begin{gathered} \mathrm{r}=0.01\mathrm{m} \\ {\mathrm{\mu }}_0=4\mathrm{\pi }\times {10}^{-7} \end{gathered}$$

Then current will be

$\mathrm{I}=5\times {10}^4\mathrm{B}$

Now for magnetic field of surface of earth, the current need to be

$$\begin{gathered} \mathrm{I}=5\times {10}^4\times 5\times {10}^{-5} \\ \mathrm{I}=2.5\mathrm{A} \end{gathered}$$

For Refrigerator magnetic field, the current needs

$$\begin{gathered} \mathrm{I}=5\times {10}^4\times 5\times {10}^{-3} \\ \mathrm{I}=250\mathrm{A} \end{gathered}$$

For laboratory magnet, the current need

$\mathrm{I}=5\times {10}^3\mathrm{to}5\times {10}^4\mathrm{A}$

For superconducting magnet, the current required

$$\begin{gathered} \mathrm{I}=5\times {10}^4\times 10 \\ \mathrm{I}=5\times {10}^5\mathrm{A} \end{gathered}$$