Question

A very long wire carrying a current I=5.0 A is bent at right angles. The magnetic induction at a point lying on a perpendicular normal to the plane of the wire drawn through the point of bending, at a distance r=25 cm from it is $2\sqrt{2}\times {10}^{-x}$ T. Find the value of x.

Answer 1

$B=\frac{{\mu }_o}{4\pi }\frac{i}{r}(sin{0}^o+sin{90}^o)=\frac{{10}^{-7}\ast 5}{25\ast {10}^{-2}}(1+0)=0.2\ast {10}^{-5}T=2\sqrt{2}\ast {10}^{-6}T$

$x=6$