# Homework Help: Another question about Magnetic fields

1. Feb 9, 2005

### andrew410

A 0.245-kg metal rod carrying a current of 10.5 A glides on two horizontal rails 0.480 m apart. What vertical magnetic field is required to keep the rod moving at a constant speed if the coefficient of kinetic friction between the rod and rails is 0.100?

Again I am supposed to use the formula F=IL x B. But I'm not sure where to start... What am I supposed to do with the coefficient of kinetic friction?

Any help would be great! thx in advance!! :)

2. Feb 9, 2005

### s_a

The coefficient of kinetic friction is simply the ratio of the frictional force acting on a body, to the normal (to the surface it moves on) component of the force acting on it (in this case simply it's weight)

So a coefficent of kinetic friction of 0.1, means that the friction force on the metal rod is a tenth of it's weight (which is easy to find). Determine the magnetic force required to cancel out this friction force (to give constant speed), and then you can determine the magnetic field required.

Last edited: Feb 9, 2005
3. Feb 9, 2005

### andrew410

So should I set the frictional force equal to the magnetic force (IL x B) and solve for B?

4. Feb 9, 2005

### s_a

Yes, you should.

5. Feb 9, 2005

### andrew410

how would I get the length though? the problem only states how far the two rails are apart from one another.

6. Feb 9, 2005

### s_a

I'd just assume the length is the distance between the rails, otherwise there wouldn't be much point in telling us the distance between the rails (actually the length of the metal rod isn't relevant anyway, only the length of it exposed to the magnetic field)

7. Feb 9, 2005

### andrew410

I don't understand what I'm doing wrong here. I set the frictional force equal to the magnetic force like this: (.1*m*g) = ILB. Then, I solve for B and get: B = (.1*m*g)/(IL). This gives me .0476 T. It says that this answer is wrong. What am I doing wrong here?

Nevermind...its fine...stupid ilrn program wasn't working properly...

Last edited: Feb 9, 2005