Magnetic Torque on Dipole; Oscillating Magnet question

[nbh]

Homework Statement

A cylindrical bar magnet whose mass is 0.08 kg, diameter is 1 cm, length is 3 cm, and whose magnetic dipole moment is <5, 0, 0> A · m2 is suspended on a low-friction pivot in a region where external coils apply a magnetic field of <1.4,0,0> T

You rotate the bar magnet slightly in the horizontal plane and release it. (For small angles in radians, assume
sin(θ) ≈ θ.)

a) What is the angular frequency of the oscillating magnet?

b) What would be the angular frequency if the applied magnetic field were <2.8,0,0> T?

Homework Equations

τ[/B] = μ × B
τ = μBsinθ
ω = qB/m
μ = IA

The Attempt at a Solution

My first instinct was to find the magnetic torque on the magnet, using the magnet's dipole moment and the applied magnetic field, but that gives zero.

 

[TJGilb]

If you had just left it sitting as is, it would be zero since the magnetic dipole moment and applied magnetic field are in the same direction. But then, the problem statement says that you rotate it in the horizontal plane which presumptively is giving one of the other axes some value. It's hard to say which without a picture of the problem or defining how the magnet is oriented, but it's probably the y component.

 

[nbh]

I figured it out. Using the moment of inertia, I, for a cylinder, you can find the angular frequency of the spinning magnet.
I = (mass*length²) /12

Solve for I and use it and μ & B to solve for ω.

ω = √(μB/I)