Question

A copper wire of diameter $1.6mm$ carries a current 𝑖. The maximum magnetic field due to this wire is $5\times {10}^{-3}T$ The value of $i$ in ampere is

Answer 1

Magnetic field due to a solid cylindrical wire increases linearly from centre (axis) to the surface and decreases from surface to infinity.
So, the magnetic field is maximum at the surface.

$B_{max}=\frac{{\mu }_{0}i}{2\pi r}$

($r=$ distance of point from axis )
(at the surface, $r=R=\text{radius}=\frac{\text{diameter}}{2}=\frac{1.6}{2}mm$

$5\times {10}^{-3}=\frac{4\pi \times {10}^{-7}\times i}{2\pi \times 0.8\times {10}^{-3}}$

$i=20A$