Magnitude of the magnetic field of a bar magnet that has a magnetic moment

1. Oct 2, 2011

MarieWynn

1. The problem statement, all variables and given/known data
A slender, 200- g bar magnet 15.0 cm long has a magnetic moment of 5.50 A m2. It is held at an angle of 45 o to the direction of a uniform magnetic field. When released, it begins to rotate about its center with an initial angular acceleration of 8.40×103 rad/s2. What is the magnitude of the magnetic field?

2. Relevant equations
T=torque
T=Ia
I=1/12mL^2
T=mu*B*sin(theta)

3. The attempt at a solution
Using the equations above, you can set Ia and mu*B*sin(theta) equal to one another and get: mu*B*sin(theta)=(mL^2/12)*a and then solve for B, which gives you:
B=(mL^2/12)*a/mu*sin(theta). Making the necessary conversions and plugging in the numbers from the question gives me:
B=(0.2*0.15^2/12)*8.4x10^3/5.5*sin(45)=0.673 T

I've gone over the problem multiple times and keep getting the same answer, which my homework is telling me is wrong. Did I miss a crucial step? Thanks!