Answer
See the detailed answer below.
Work Step by Step
We know that the magnetic field of a long straight current wire is given by
$$B=\dfrac{\mu_0I}{2\pi d}$$
Hence, the distance needed from the long thin wire is given by
$$d=\dfrac{\mu_0I}{2\pi B}$$
For current of $I=10\;\rm A$;
$$d=\dfrac{(4\pi \times 10^{-7})(10)}{2\pi B}$$
$$d=\dfrac{(20 \times 10^{-7}) }{ B}$$
Using the data in Table 32.1, to find the distances needed,
$$d_1=\dfrac{(20 \times 10^{-7}) }{ B_{\rm Earth's \;surface}}=\dfrac{(20 \times 10^{-7}) }{ (5\times 10^{-5})}=\color{red}{\bf 4}\;\rm cm$$
$$d_2=\dfrac{(20 \times 10^{-7}) }{ B_{\rm Refrigerator\;magnet}}=\dfrac{(20 \times 10^{-7}) }{ (5\times 10^{-3})}=\color{red}{\bf 0.04}\;\rm cm$$
$$(d_3)_{\rm minimun}=\dfrac{(20 \times 10^{-7}) }{ B_{\rm Lab\;magnet}}=\dfrac{(20 \times 10^{-7}) }{ (0.1)}=\color{red}{\bf 20}\;\rm \mu m$$
$$(d_3)_{\rm maximum}=\dfrac{(20 \times 10^{-7}) }{ B_{\rm Lab\;magnet}}=\dfrac{(20 \times 10^{-7}) }{ (1)}=\color{red}{\bf 2.0}\;\rm \mu m$$
$$d_4 =\dfrac{(20 \times 10^{-7}) }{ B_{\rm Lab\;magnet}}=\dfrac{(20 \times 10^{-7}) }{ (10)}=\color{red}{\bf 0.2}\;\rm \mu m$$
