Question

A student makes a short electromagnet by winding 580 turns of wire around a wooden cylinder of diameter d = 2.1 cm. The coil is connected to a battery producing a current of 5.1 A in the wire. (a) What is the magnitude of the magnetic dipole moment of this device? (b) At what axial distance z >> d will the magnetic field have the magnitude 4.1 µT (approximately one-tenth that of Earth's magnetic field)?

Answer 1

Answer:

1.02453 Am²

0.36834 m

Explanation:

N = Number of turns = 580

d = Diameter = 2.1 cm

r = Radius = $\frac{d}{2}=1.05\ cm$

A = Area = $\pi r^2$

I = Current = 5.1 A

B = Magnetic field = 4.1 μT

x = Distance

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

Magnetic dipole moment is given by

$\mu=NIA\\\Rightarrow \mu=580\times 5.1\times \pi\times 0.0105^2\\\Rightarrow \mu=1.02453\ Am^2$

The magnitude of the magnetic dipole moment of this device is 1.02453 Am²

Magnetic field is given by

$B=\frac{\mu_0NIA}{2\pi x^3}\\\Rightarrow x=\left(\frac{\mu_0NIA}{2\pi B}\right)^{\frac{1}{3}}\\\Rightarrow x=\left(\frac{4\pi\times 10^{-7}\times 580\times 5.1\times \pi\times 0.0105^2}{2\pi 4.1\times 10^{-6}}\right)^{\frac{1}{3}}\\\Rightarrow x=0.36834\ m$

The axial distance is 0.36834 m