# Finding Magnetic Field

1. Jul 9, 2008

### gtqueen371

1. The problem statement, all variables and given/known data

A metal rod of length 14 cm and mass 80 grams has metal loops at both ends, which go around two metal poles. The rod is in good electrical contact with the poles but can slide freely up and down. A magnet supplies a large uniform magnetic field B in the region of the rod.

When the rod is not connected to a battery, and no current runs through it, if it is released in the position shown in the diagram, it falls downward until it hits the table.
However, when the metal poles are connected by wires to a battery, and a 3 ampere conventional current flows through the rod in the direction shown, the rod remains at rest when released in the position shown, and does not fall.

In the diagram +x is to the right, +y is up, and +z is out of the page.

What is the magnitude of the magnetic field due to the large magnet, at the location of the rod? (The magnetic fields due to the current in the current in the circuit are negligible compared to the magnetic field of the magnet).
Bmagnitude =_______ Tesla

What is the direction of the magnetic field due to the large magnet, at the location of the rod? _________

2. Relevant equations

Fmagnetic = I$$\Delta$$L x B

eE = qvB

F = qv x B

I = qnAv

3. The attempt at a solution

I don't even really know where to start. I feel like I don't have enough information. I have I, L and weight. If v = I/qnA, I plug this into F = q(I/qnA) x B = (I/nA) x B but I still don't know n or A...

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2. Jul 9, 2008

### Staff: Mentor

You'll need this formula for the magnetic force on the current-carrying wire. (You'll also need the associated right hand rule to find the direction.)

What forces act on the rod? Since the rod remains at rest, what must the net force be?

3. Jul 9, 2008

### gtqueen371

The net force must equal zero. So F = qE + qvB becomes -qE = qvB which then is E=vB but I don't have E or v. I don't understand how this fits together with Fmagnetic = I$$\Delta$$L x B...

Also if I use Fmagnetic = I$$\Delta$$L x B with -qE = qVB, then I get -qE = I$$\Delta$$L x B which then equals (qE)/(I$$\Delta$$L) = B, but now I am missing E...

4. Jul 9, 2008

### Staff: Mentor

There's no E, only B, so forget about qE. There are two forces acting on the rod. What are they?

5. Jul 9, 2008

### gtqueen371

gravitational and magnetic

6. Jul 9, 2008

### Staff: Mentor

Right! Gravity acts down so the magnetic force better act up.

Set up an equation between those two forces. Then you can solve for B.

7. Jul 9, 2008

### gtqueen371

mg = I$$\Delta$$L x B
(.08kg)(9.81 m/s/s) = (3A)(.14m) x B
1.869 T = B in the -z direction!!!