Question

calculate the magnitude of the magnetic field at a point 25.00 cm from a long thin conductor carrying a current of 2.00 A

Answer 1

The magnitude of the magnetic field is, \(\begin{aligned} & \oint \vec{B} \cdot d \vec{s}=\mu_{0} I \\ B .2 \pi r=& \mu_{0} I \\ B=& \frac{\mu_{0} I}{2 \pi r} \\=& \frac{\left(4 \pi \times 10^{-7}\right)(2)}{2 \pi(0.25)} \\=& 1.6 \times 10^{-6} \mathrm{~T} \end{aligned}\)