- #1

jm321710

- 1

- 0

## Homework Statement

I am so stuck on this, and would be very thankful for any help!

a). Calculate the magnetic dipole moment of a single atom, based on the following model: One electron travels at speed 7.40 × 10^6 m/s in a circular orbit of diameter 2 × 10^-10 m.

b). The individual atomic magnetic dipoles of magnetic materials (such as iron) are preferentially lined up to point in the same direction. If a fraction f of the dipoles are so aligned along the long axis (with the rest oriented randomly so that their magnetic dipole moments add vectorially to zero), what is the net magnetic dipole moment of a piece of such material (as shown in the diagram below); where w = 5.15 cm, h = 7.15 cm, l = 10.0 cm, and f = 0.40? Note, the material may be viewed as an array of cubes, each of which contains one atom, with only the outermost electron being considered, and is 2 × 10^-10 m on a side. Also assume the same speed for electron as in part a.)

c).What is the torque experienced by the piece of material in a field of 4.30 × 10-3 T, when the magnetic field is directed at right angles to the long axis of the material, as shown in the diagram?

## Homework Equations

M=NIA

i = Q/t

v= Δx/Δt

## The Attempt at a Solution

For part a). Area of the electron: A = π(1x10^-10)^2 = 3.4 x 10^-20

N = (7.4x10^6)/(2x10^-10) = 3.7 x 10^16

t = Δx/v => (2x10^-10)/(7.4x10^6) = 2.7 x 10 ^-17

i = (-1.6x10^-19)/(2.7 x 10 ^-17) =-0.005962

M = (3.7 x 10^16)(-0.005962)(3.4 x 10^-20)

The right answer should be 5.93 x 10 ^ -23, but I am doing something very wrong. I would be so thankful for any help.