What Is the Minimum Magnetic Field Strength to Move a Rod in a Magnetic Field?

[Dopefish1337]

Homework Statement

Suppose the rod in the figure below has mass m = 0.375 kg and length 18.5 cm and the current through it is I = 36.2 A.

http://img183.imageshack.us/img183/2059/dgian2757.gif

If the coefficient of static friction is mu_s = 0.503, determine the minimum magnetic field B (not necessarily vertical) that will just cause the rod to slide. Give the magnitude of B and its direction (angle) relative to the vertical.

Homework Equations

|F|=I*l*B*sin(theta)
F_f=F_n*mu_s
F_n=m*g

The Attempt at a Solution

F_n=3.67875 N
F_f=1.85 N

The magnetic force needs to be strong enough to overcome F_f, so F_f=I*l*B*sin(theta). Surely that force is maximised when sin theta is one, so relative to the vertical that angle must be 0?

Thus, 1.85=36.2*(18.5/100)*B, so B= .276 T, at an angle of 0 deg relative to the horizontal, however this is incorrect.

At what point am I going wrong?

 

[Delphi51]

Looks good to me! It does ask for the angle relative to the vertical (rather than horizontal as you mention in the 2nd last paragraph) but that is zero so 0.276 and zero should be the answer.

 

[Dopefish1337]

Unfortunately the system isn't accepting it. Anyone else want to look for any reason it might be wrong, before I email the admin for the site claiming it's a technical error?

 

[Dopefish1337]

I'm toying with the idea that perhaps it might be possible that if the field were on a slight angle, the rod itself might feel a slight vertical force, thus causing the normal force to decrease and in turn the frictional force.

However, I don't think that would actually decrease the field needed, since overcoming the frictional force is easier than the gravatational one...

Anyone with a better understanding of all this want to comment on whether it's possible for an angle to be better than it going straight vertical?

 

[Delphi51]

A most impressive idea!
I graphed the part that depends on the angle and there seems to be a minimum for B at about 27 degrees away from vertical.

 

[Dopefish1337]

What would the resultant equations be that lead to that conclusion?

F_n=mg-I*l*b*cos(theta)?

So that I*l*b*sin(theta)=(mg-I*l*b*cos(theta))*.503?

Or did I mess up my expression for F_n?

 

[Dopefish1337]

Got it now, think my cos and sin were reversed in my previous post, but I fiddled with things until it worked, and it has.

Thanks for the help. :)

 

[dancer123]

I am working on a similar problem, but cannot find the solution.
I have gotten to finding an equation:

Force Friction = I l B sin(theta)
But cannot solve for Theta and B
Help?

 

[Dopefish1337]

Force friction = I l B sin(theta)

What expression do you have for frictional force? (or are you saying that the frictional force *is* ILBsin(theta), as opposed to simply be equivalent?)

Once you have an expression with only B and theta as unknowns, you can rearrange to get B as a function of theta. Once you have that, there's two possibilities: 1. Find a graphing calcluator, and look for the point where B is minimum directly off the graph.

or

2. Draw upon calculus to find the derivative and set it equal to zero, and solve the resulting equation for theta.

If memory serves, you may not have mention of theta in the numerator, so it may be easier to just take the denominator and maximise that, potentially leading to an easier derivative to solve.

 

[dancer123]

I have used Frictional Force to equal Normal force * mew(Static)
the question never stated in which direction the rod was hung, so I neglected any angle from the frictional force.
I have a graphing calculator, but how would I solve for it?
Would i simply divide my calculated value by sin(theta)?

 

[Dopefish1337]

I suspect if your question is similar to mine, then you'd want an angle in your formula for normal force.

If you plot the function of B as a function of theta, you can just manually look through the table and find where B is minimum.