Oscillation of a coil in a Magnetic Field

[mydir]

Homework Statement

Hi I am really stuck on this question. It got handed out to us this morning and is due in Wednesday of this week. Any help would be great thanks.

A circular solid coil consisting of a circular disk of radius R = 20mm and mass M =
50 g pivoted through its diameter. It has 10 loop of thin wire round it circumference carry a current i = 400mA.
This coil is placed in a constant magnetic field of B = 100mT orientated such that the
torque on the coil is a minumum.
Show that if the coil is rotated by a small angle and released, it will undergo simple
harmonic motion about the minimum torque position. Calcualte the frequency of the
oscillation.

Homework Equations

Torque =μ(magnetic monent) ×B [Vector equation]
F(magnetic) = qv B
F(central) =(mv^2)/r
Torque= AIB sin(theta) [A is the area of the loop]

The Attempt at a Solution

I don't really know where to start. When it says the torque is a minimum does that its 0?
Should i solve the two force equations for r: r =mv/q B? Then solve for the frequency w=qB/m? Could the moment of inertia I = (1/4)mr^2, help me in this question? I mean I really don't know where to start and any help whatsoever would be much appreciated.

Thanks
mydir

 

[brainstorm39]

hi dymir,

We know that the restoring torque of this system is:
τ=-NAIBsin(θ), (where N is the number of turns in the coil)

This means,
I_M*a ̈=-NAIBsin(θ), (where I_M is the moment of inertia, and a is the angular acceleration)
So for small angles:
I_M*a ̈=-NAIBθ,
Hence the EOM is of the form,
a+(ω^2)*θ, (where ω^2=NAIB/I_M)
This means there is harmonic motion,

To find the frequency of oscillation, we have angular frequency ω,

So the frequency is just;
f=ω/2pi

Now u have the method, You can do the math^^

 

[mydir]

Thank you very much for the help. I had spotted thta torque was alsoe I_M*a but I hadnt realized that w^2 = NAIB/I_M. Now however all is fine.

Thanks again brainstorm!

mydir