Torque on a coil in a magnetic field

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SUMMARY

The discussion centers on calculating the minimum current required to prevent a cylinder from rolling down an inclined plane in a magnetic field. The cylinder has a mass of 262 g, a length of 12.7 cm, and is wound with 13 turns of wire. Using the torque equation τ = NiABsin(θ) and balancing it with the gravitational torque τ_g = mgrsin(θ), the minimum current is calculated to be 1.63 A. The discussion also addresses concerns about the role of friction and the consistency of torque reference points in the solution.

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  • Understanding of torque equations in physics
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  • Knowledge of inclined plane dynamics and forces
  • Basic algebra for solving equations
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Homework Statement


A cylinder with mass m = 262 g and length L = 12.7 cm has a wire longitudinally rolled up around it, such that the rolling plane, with N = 13 turns, contains the axis of the cylinder and is parallel to a plane which has an inclination θ with the horizontal. The cylinder is on this inclined plane, and the set is under an uniform magnetic field, of 477 mT. What is the smallest current which must go through the coil, such that the cylinder doesn't roll down?
See the attached picture for more information [it is scanned from "Physics", Vol. 3, by Halliday, Resnick and Krane, Brazilian edition].

Homework Equations


Torque on a coil with N turns (or loops):
\tau = NiAB\sin{\theta}
where i is the current flowing through the coil, A is the area of one loop, B is the magnitude of the magnetic field and θ is the angle between the vector normal to the coil's plane and the magnetic field vector.

The Attempt at a Solution


The intersection between the cylinder and the inclined plane is a line. The torque due to gravity with respect to the contact points is:
\tau_g = mgr\sin\theta
With respect to the center of mass of the cylinder, the torque due to the magnetic field, when there is a current i flowing through the coil, is:
\tau_B = NiAB\sin\theta = Ni(2rL)B\sin\theta
and tends to make the cylinder roll up.
For the net torque to be zero, the torques due to the magnetic field and gravity must have equal magnitudes:
mgr\sin\theta = Ni(2rL)B\sin\theta
mg = 2NiLB
i = \frac{mg}{2NLB}
Plugging in the values, I obtain i = 1.63 A, which is the book's answer.
However, I'm not very sure about this solution. My doubts are:
1) It appears that there shouldn't be any rolling in the first place, since there is no mention of friction. I thought that friction with the inclined plane, and not weight, would be the cause of rolling. Weight and no friction would cause the cylinder to simply slide.
2) I used two different reference points in order to measure the torques. Nevertheless, I got to the answer by equaling their values. But this doesn't seem very consistent; why does it work? In order to be equaled, shouldn't the torques be measured with respect to the same reference point?

Thank you in advance.
 

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While you are right that friction is required for rolling I think you will have to let this little error in the question go.
 

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