The magnetic field at a distance of 2 cm from a long straight wire is \(1\times10^{-4}\ \mathrm{T}\). The current in the wire is now doubled. The magnetic field at a distance of 8 cm from the wire is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(5.0\times10^{-5}\ \mathrm{T}\)
For a long straight wire, \( B=\mu_0I/2\pi r \). So \( B \) is directly proportional to \( I \) and inversely proportional to \( r \). The ratio of the two fields needs only these two changes.
\( \begin{aligned} \frac{B_2}{B_1} &= \frac{I_2}{I_1}\times\frac{r_1}{r_2} \\ &= 2\times\frac{2}{8} \\ &= \frac{1}{2} \end{aligned} \)
\( \begin{aligned} B_2 &= \frac{1}{2}\times1\times10^{-4} \\ &= 5.0\times10^{-5}\ \mathrm{T} \end{aligned} \)
NCERT (2026-27):
Ampere’s Circuital Law – “… The field is directly proportional to the current and inversely proportional to the distance from the (infinitely long) current source.”