Answer:
1.02453 Am²
0.36834 m
Explanation:
N = Number of turns = 580
d = Diameter = 2.1 cm
r = Radius = [tex]\frac{d}{2}=1.05\ cm[/tex]
A = Area = [tex]\pi r^2[/tex]
I = Current = 5.1 A
B = Magnetic field = 4.1 μT
x = Distance
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]
Magnetic dipole moment is given by
[tex]\mu=NIA\\\Rightarrow \mu=580\times 5.1\times \pi\times 0.0105^2\\\Rightarrow \mu=1.02453\ Am^2[/tex]
The magnitude of the magnetic dipole moment of this device is 1.02453 Am²
Magnetic field is given by
[tex]B=\frac{\mu_0NIA}{2\pi x^3}\\\Rightarrow x=\left(\frac{\mu_0NIA}{2\pi B}\right)^{\frac{1}{3}}\\\Rightarrow x=\left(\frac{4\pi\times 10^{-7}\times 580\times 5.1\times \pi\times 0.0105^2}{2\pi 4.1\times 10^{-6}}\right)^{\frac{1}{3}}\\\Rightarrow x=0.36834\ m[/tex]
The axial distance is 0.36834 m
