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

1.02453 Am²

0.36834 m

Step-by-step explanation:

N = Number of turns = 580

d = Diameter = 2.1 cm

r = Radius =
`(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 * 10^(-7)\ H/m`

Magnetic dipole moment is given by

`\mu=NIA\\\Rightarrow \mu=580* 5.1* \pi* 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=(\mu_0NIA)/(2\pi x^3)\\\Rightarrow x=\left((\mu_0NIA)/(2\pi B)\right)^{(1)/(3)}\\\Rightarrow x=\left((4\pi* 10^(-7)* 580* 5.1* \pi* 0.0105^2)/(2\pi 4.1* 10^(-6))\right)^{(1)/(3)}\\\Rightarrow x=0.36834\ m`

The axial distance is 0.36834 m