What Is the Minimum Magnetic Field to Move a Copper Rod?

[Uku]

Homework Statement

Right, this is from Halliday & Resnick, ch. 28 p45
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A 1kg copper rod rests on two horizontal rails 1m apart and carries a current of 50A from one rail to the other. The coefficient of static friction between the rod and rails is 0.6
What are the a) magnitude and b) angle (relative to the vertical) of the smallest magnetic field that puts the rod on the verge of sliding?
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Answers being a) B=0.1T and b)31deg

The Attempt at a Solution

The force generated by the magnetic field on the bar has to be equal to the normal force times the coefficient of s. friction. This is:

$$iLxB=mg\mu$$

which can be written as

$$iLBsin\alpha=mg\mu$$

We can write this as

$$Bsin\alpha=\frac{mg\mu}{iL}$$

Clearly, the right hand side is constant, independent of the angle. So, in order for example for the angle to change into something smaller than perpendicular (90), B must go up...

Besides, using the answers from the book I get $$F_{B}\approx3N$$
whilst $$mg\mu\approx6N$$

Does not make sense.

And to complicate things. There are two degrees of freedom in the vertical plane for the vector to rotate. One of them describes the tilting of the magnetic field toward the wire-crossed bar, this actually changes the magnitude of the force on the wire.
The other angle, which describes the tilting of the magnetic field toward the current carriers, does not change the magnitude of the force, but the projection of it onto, let's say, the x axis.

In any case, B times whatever angle must be a constant, since the force it meets is a constant. I don't see where the 31 degrees comes from.

If I assume sin90 then the field is B=0.12T

 

[AvroArrow]

, which is close to the answer. But this does not take into account the angle of the magnetic field but the angle between the rod and rails.Any help?