A straight wire of diameter $0.4 \mathrm{~mm}$ carrying a current of $2 \mathrm{~A}$ is replaced by another…
A straight wire of diameter $0.4 \mathrm{~mm}$ carrying a current of $2 \mathrm{~A}$ is replaced by another wire of $0.8 \mathrm{~mm}$ diameter carrying the same current. The magnetic field at distance (R) from both the wires is ' $\mathrm{B}_1$ ' and ' $\mathrm{B}_2$ ' respectively. The relation between $\mathrm{B}_1$ and $\mathrm{B}_2$ is
$\mathrm{B}_1=\frac{\mathrm{B}_2}{2}$
$\mathrm{B}_1=\mathrm{B}_2$
$\mathrm{B}_1=2 \mathrm{~B}_2$
$\mathrm{B}_1=\frac{\mathrm{B}_2}{3}$
Solution
Since the same current flows through the second wire, the magnetic field at the same distance will be same.
$\therefore \mathrm{B}_1=\mathrm{B}_2$