Torque on a Circular Current Loop under a Misaligned Magnetic Field

[domephilis]

Homework Statement

The text (R. Shankar's Fundamentals of Physics II) has given the derivation for a rectangular loop showing that the torque is equal to $\vec{\tau} = \vec{\mu} \times \vec{B}$. In a problem, I need to calculate the torque on a circular loop. The loop lies in the x-y plane. The magnetic field is at an angle $\theta$ to the area vector (perpendicular to the area). We need to find the torque on the loop which serves to align the area vector with the B field. We assume here that the forces add up to zero.

Relevant Equations

$$\tau = \int{(\vec{I} \times \vec{B} dl) \times \vec{r}}$$

My final result was $$\tau = IBR\sin{\theta} \int_{0}^{2\pi} {\frac{\sqrt{1+\sin^2{\theta}\cos^2{\gamma}}}{\sqrt{1+\sin^2{\theta}\cos^2{\gamma}}}d\gamma}$$. I think I am supposed to get a simple answer like $$\tau = \vec{\mu} \times \vec{B}$$ where mu=IA. If I take approximations using the Taylor Series, I can get close (twice the above written value). I will write my derivation below and please let me know where I am wrong:

The difficulty in this problem comes from finding the angles for the cross products. Note that in the below derivation $\theta$ is the angle between the area vector and the field; and $\gamma$ is the angle between the x axis and the radial vector to the infinitesimal segment of the loop under consideration (counterclockwise is positive).

Firstly for the torque on an infinitesimal segment of the loop dl, $\vec{I} \times \vec{B}$ is perpendicular to I and B. Since the x axis is perpendicular to the area vector, the force is at an angle $\theta$ to the x-axis (as B is to A). Hence, whatever the force is in magnitude, it is $\theta$ from the radial axis. Remember that for when we take the cross product later.

Secondly, consider the angle between the current and the magnetic field. I will attach in this thread a drawing I made on the whiteboard. The goal is to add theta and gamma in the correct way to get the angle between the current (perpendicular to the radial vector in the x-y plane) and the field (at angle theta to the area vector). That angle phi is such that $$\sin(\phi) = \frac{\sqrt{1+\sin^2{\theta}\cos^2{\gamma}}}{\sqrt{1+2\sin^2{\theta}\cos^2{\gamma}}}$$.

If the above two paragraphs are correct, than the formula at the beginning is correct. The rest is just a routine application of the torque formula and the Lorentz force formula. If so, does anyone know a closed form for the integral?

 

[domephilis]

Here’s the image. The lower left figure is for the second bit. The lower right one is for the torque angle bit. The lowest figure is to figure out how to add angles in 3D. As is the equations on the left side of the board.