Answer:
The radius is [tex]R=8.86mm[/tex]
Explanation:
Generally Magnetic field due to a current carrying wire is is mathematically represented as
[tex]B = \frac{\mu_0 I }{2 \pi r_1}[/tex]
Where [tex]r_1[/tex] is the distance from the axis of the wire to the first point we are considering
I is the current
[tex]\mu_0[/tex] is the permeability of free space
Making I the subject
[tex]I = \frac{2 \pi r_1 B}{\mu_0}[/tex]
Generally Magnetic field inside a current carrying wire is is mathematically represented as
[tex]B_0=\frac{\mu_0 I r}{2 \pi R^2}[/tex]
Now the r here is the distance from the axis of the wire to the second point we are considering
And R i the radius of the wire
Making R the subject of the formula
[tex]R = \sqrt{\frac{\mu_0 I r}{2 \pi B_0} }[/tex]
Substituting for I we have
[tex]R = \sqrt{\frac{\mu_0 (\frac{2 \pi r_1 B}{\mu_0} ) r}{2 \pi B_0} }[/tex]
[tex]R = \sqrt{\frac{B}{B_0} r * r_1}[/tex]
Where B = 0.27mT
And [tex]B_0[/tex] = 0.44mT
r = 32mm
[tex]r_1[/tex] = 4.0mm
Substituting values
[tex]R = \sqrt{\frac{0.27}{0.44} *4 *32 }[/tex]
[tex]R=8.86mm[/tex]
