- #1
domephilis
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- 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?
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?
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