# Currents and magnetic fields

1. Oct 28, 2006

### -_-'

question: Use the following quantities to calculate the magnetic feild strength of the horseshoe magnet.

Mass of coil: 20 grams
current in coil: 2.0 amps
length of bottom of coil: 10cm
angle coil makes with vertical: 30 degrees

Hint: Equate magnetic force on bottom of coil to the horizontal componen of the coils weight.

I've got a diagram
http://i122.photobucket.com/albums/o272/science_f/0000000000000coilandmagnet.jpg

I don't know where to start in the calculating proccess Can someone please help me with the starting point and i'll try and progress from there....this is the first time i've seen a problem like this

2. Oct 28, 2006

### -_-'

these are the formulas I know:
B=F/IL
F=BIL sin @
B=kI/r
B=2(pi)knI

Where k is Amperes constant (2x10^-7NA^-2), L is the length of wire, I is the current, r is the radius of the wire (but in some cases the distance between wires), and F is the force.

Here's another diagram with brainstorming but i'm not sure if i grasp the concept of this.....I think it is all wrong

http://i122.photobucket.com/albums/o272/science_f/000000000000h.jpg

I actually think that there isn't enough information to do the problem unless it is as simple as:
F=mg=20x9.8=196
B=F/IL
B= 196/2.0x10
B=9.8
But i don't think that can be right

Last edited: Oct 28, 2006
3. Oct 28, 2006

### OlderDan

What is a horizontal component of weight? I don't think they really mean that. The magnet will exert a force on the lower branch of the coil. You need to think about the direction of the magnetic force. You also need to think about what is keeping the coil in position. There has to be something besides the magnetic force. Something has to be keeping the top of the coil in position. What will be the direction and magnitude of the force on the top of the coil? It will help if you rotate the image so that the magnet is in the plane of the page, and the loop is perpendicular to the page so that the 30 degree tilt can be seen.

Last edited: Oct 28, 2006
4. Oct 28, 2006

### -_-'

thanks for that every little bit of info helps

5. Oct 29, 2006

### OlderDan

Hopefully you will recognize that this becomes a statics problem where one of the forces involved is a magnetic force. Think forces and torques and all the usual things you use for statics problems.