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

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A steel rod weighing 0.10kg is placed on two inclined metal rails at a 30-degree angle, within a magnetic field of 0.50T and carrying a current of 70A. The force due to the magnetic field acting on the rod is calculated using the formula Fb = BIL, resulting in a force of approximately 3.64N directed into the page. The normal force (Fn) is determined to be 0.8487N, while the gravitational force component along the incline is calculated as mgsin30. The net force acting on the rod along the incline is found by subtracting the gravitational force from the magnetic force, yielding a resultant force of 3.71N up the slope. The discussion concludes that the problem resembles a box on an incline scenario, with the magnetic force acting to the right up the ramp.
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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!
 
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anyone ?
 


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
 


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

Fb = BIL
 


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?
 


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.
 


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?
 


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.
 

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