# Magnetic field strength needed to lift a heavy conductor

1. Apr 13, 2008

### kheila

1. The problem statement, all variables and given/known data
A heavy conductor (mass m, length l, resistance R) is suspended by two springs each with spring constant k, and connected to a battery with electric potential V as shown in the figure. A magnetic field B is now imposed. The acceleration of gravity is 9.8 m/s^2.

I'm going to describe this image as best as I can, as the site won't let me link to an outside image: There is a battery with a wire coming out horizontally from each side. The positive side of the battery is on the left. There is a spring hanging down from the end of each wire. Attached to the bottom of the springs is a long rod, with a spring attached to either end, allowing it to hang parallel to the ground.

What is the minimum magnetic field strength required to completely take the weight of the heavy conductor off the springs if the potential voltage is 20V, conductor resistance is 11 ohms, length is .63 m, spring constant is 1.7 N/m, conductor mass is 1.6 kg?

2. Relevant equations
I=R/V
F=ma
F(spring) = -kx
F(b) = I*L x B (cross product)
where L is the vector in the direction of I w/ magnitude of the conductor length l and F(b) is the resultant force exerted by a magnetic field on a current-carrying wire (I assume this can also be applied to a current-carrying conductor)

3. The attempt at a solution
I drew a free body diagram for the conductor. The force downward is mg. The forces upward and opposite mg are 2kx (that is: F(net spring) = k1*x + k2*x, and k1 and k2 are equal), and I*L x B.

This is my reasoning: in order for the weight to be completely taken off of the springs, the upward force F(b) must equal the downward force, which is mg. Therefore, the spring force does not need to be taken into consideration.

mg = I*L x B
mg = (I * L-sub-x * i-hat) x (B-sub-z * k-hat) <--since the only value of L is in the y-direction, and the only value of B is in the z-direction

mg = -I * L-sub-x * B-sub-z * j-hat
B-sub-z * j-hat = -mg/(I*L-sub-x)

Plugging in the numbers gives me -45.252525...

Am I reasoning the FBD correctly? Am I right that I should just set F(b) equal to mg?

2. Apr 13, 2008

### G01

Your reasoning is correct, but I think you are making the problem harder than it is.

HINT: Try working just with magnitudes. Assume the B-field points in the direction needed so that the magnetic force points upward. Working with magnitudes will make your starting equation equating the two forces equal to the following:

$$mg=ILB\sin\theta$$ where theta is the angle between the L and B vectors.

Now, we want the minimum B field. What does this mean for any terms in the above equation?

3. Apr 13, 2008

### kheila

Doesn't theta = 90 degrees, so the sin(theta) term can be dropped? Also, this is the second part of a two-part problem. In the first part, it was established that the magnetic field is perpendicular to the "paper."

I see how the equation can be simplified, as you showed. However, I don't understand how the equation would change in order to find the "minimum" B field. Wouldn't just solving for B give that automatically?

Last edited: Apr 13, 2008
4. Apr 13, 2008

### G01

Yes theta is 90 degrees given your description of the magnetic field perpendicular to the paper. What they must mean by minimum B-field is that the magnetic force is "just" canceling with the weight force. In other words, the magnetic force is exactly equal to the weight, not any greater. This will happen if we use the minimum B-Field that will take the force off of the springs.

So, the assumptions you made have already made sure that the B-field you get is the minimum B-field.

5. Apr 13, 2008

### kheila

Ok. So, solving that equation for B:

B = mg/(I*L)
= (1.6 kg)*(9.8 m/s^2) / (I*.63 m)
I = R/V = 11 ohms/20V = .55 A
Plug this in: B = (1.6 kg)*(9.8 m/s^2)/(.55A*.63m)
= 45.25252525... T

But this is not the right answer. And that's almost exactly what I did the first time.

6. Apr 13, 2008

### G01

I=V/R, not R/V

I think that is your problem. Didn't catch that error at first, sorry.

7. Apr 13, 2008

### kheila

Agh!!

Thanks for helping me through this :)