Question

please help me to solved this Compute the flux density in air at a point 6.0 cm from a long straight wire carrying a current of 9.0 A.

Answer 1

magnetic flue density=fluex/unit area

fluex=B. A.N

fluex density=B.A.N/A

=B (no of turns 1)

magnetic field produced by long conducting wire= u.i/2πr

B= 4π×10^-7×9/2π×6×10^-2

fluex density =3×10^-5 T

Answer 2

We know that the magnetic field strength or magnetic flux density produced by a long straight current-carrying wire equals 

$B=\frac{\mu_{0}I }{2\pi r}$

where,

 $I$ is the current,

 $r$ is the shortest distance to the wire and,

$\mu_{0}$  is the permeability of open space

Given data,

$I=9A$

$r=6cm$

$\mu_{0}=4\pi\times10^{-7} m/A$

$1cm=10^{-2}m$

To find the flux density,

$B=\frac{4\pi\times10^{-7} \times9}{2\pi\times6\times10^{-2} } =3\times10^{-7+2}$

$B=3\times10^{-5} T$

Hence, the flux density produced by the long straight wire is $3\times10^{-5} T$