A long straight wire carries a current of \( \pi \) A. At what distance from the wire is the magnetic field due to it equal to \( 5\times10^{-5}\,\mathrm{T} \)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 10^{4}\mu_0\,\mathrm{m} \)
The magnetic field of a long straight wire carrying current \( I \) at a distance \( r \) from it is \( B=\mu_0I/2\pi r \). So the field has a given value at the distance \( r=\mu_0I/2\pi B \).
With \( I=\pi\,\mathrm{A} \), \( B=5\times10^{-5}\,\mathrm{T} \) and \( \mu_0 \) in SI units:
\( \begin{aligned} r &= \frac{\mu_0I}{2\pi B} \\ &= \frac{\mu_0\times\pi}{2\pi\times5\times10^{-5}}\,\mathrm{m} \\ &= \frac{\mu_0}{10^{-4}}\,\mathrm{m} \\ &= 10^{4}\mu_0\,\mathrm{m} \end{aligned} \)