Question

A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Answer 1

Current in the wire, \(I = 35 A\)
Distance of a point from the wire, \(r = 20\,cm = 0.2\,m\)
Magnitude of the magnetic field at this point is given as:
                           \(|B|= \frac{μ_0}{4π}. \frac{2l}{r}\)
Where,
\(μ_0\)= Permeability of free space
    \(= 4π × 10^{–7} T m A^{–1}\)
So, 
                 \(|B| =\frac{4π × 10^{–7}}{4π} × \frac{2 × 35}{0.2}\)
                       \(= 3.5× 10^{-5}\, T\)
Hence, the magnitude of the magnetic field at a point 20 cm from the wire is \(3.5× 10^{-5}\, T\).