# A steel rod of 0.10kg rests on two metal rails inclined at an angle of

1. Jan 15, 2012

### SpecialKM

A steel rod of 0.10kg rests on two metal rails inclined at an angle of 30 degrees. There is a magnetic field of 0.50T on the incline and a current of 70A flowing across the bar. Find the acceleration of the bar.
The bar is 0.12m wide**

The attempt at a solution

What I have so far is a diagram of the problem, but I'm not sure if that is correct.
My intuition is that you'd need to make an fbd of all the forces. On my fbd I have:
Fn, Fg and Fb (force due to the magnetic field). I know only a portion of the 0.50T interacts
with the rod. I'm just confused where I put the Fb in my sum of forces statements. I have two,
Fnet (paralell) and Fnet(perpendicular) [to the motion]. Thanks!

Last edited: Jan 15, 2012
2. Jan 15, 2012

### SpecialKM

Re: Electromagnetism

anyone ?

3. Jan 16, 2012

### technician

Re: Electromagnetism

I am not certain of the physical arrangement without seeing a diagram but the main equation you need gives the force on a conductor carrying a current in a magnetic field
Do you know this equation

4. Jan 16, 2012

### SpecialKM

Re: Electromagnetism

well, what I know is the force of a wire interacting with an external B field.

Fb = BIL

5. Jan 16, 2012

### technician

Re: Electromagnetism

That is correct for a field at 90 to the rod. The 30 degrees mentioned in your question must have some relevance. Do you have a diagram
What value of force do you get using the numbers in the question?

6. Jan 16, 2012

### SpecialKM

Re: Electromagnetism

Well this is what I get:

Fb = BIL(perpendicular)
Fb = 0.5T(70A)(0.12mcos30)
Fb = 3.6373N into the page

Fnet (perpendicular) = Fn - mgcos30

Fn = mgcos30
Fn = (0.1kg)(9.8 N/kg)cos30
Fn = 0.8487N

There's also mgsin30, but I don't know where that goes. This is where I'm stuck.

7. Jan 16, 2012

### technician

Re: Electromagnetism

I still can't picture the arrangement.....if the 2 rails are at 30 to the horizontal and the magnetic field causes a force up the slope of BIL = 4.2N then the resultant force up the slope = 4.2 - (0.1 x 9.81 x Sin30) = 4.2 - 0.49 = 3.71N
does this fit with your arrangement?

8. Jan 17, 2012

### SpecialKM

Re: Electromagnetism

I think I've figured it out, It's just like a box on an incline question, however the current is going out of page, resulting in a magnetic force to the right up the ramp.