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

In summary, a 0.10kg steel rod resting on two metal rails inclined at 30 degrees experiences a resultant force of 3.71N up the ramp due to a magnetic field of 0.50T and a current of 70A flowing across the bar. This can be calculated using the equation Fb = BIL and taking into account the forces of gravity and the normal force. The magnetic force is directed to the right up the ramp due to the current going out of the page.
  • #1
SpecialKM
52
0
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:
Physics news on Phys.org
  • #2


anyone ?
 
  • #3


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


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

Fb = BIL
 
  • #5


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


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


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


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.
 

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

1. What is the weight of the steel rod?

The weight of the steel rod is 0.10kg, as stated in the question.

2. How are the metal rails inclined?

The metal rails are inclined at an angle, but the exact angle is not specified in the question.

3. What is the purpose of the metal rails?

The metal rails are likely being used to support or hold the steel rod in place, as they are mentioned in relation to the rod.

4. Is there any other information about the steel rod?

No, the question only mentions the weight of the steel rod and its placement on the metal rails.

5. What are the implications of the steel rod resting on inclined metal rails?

The inclination of the metal rails may affect the stability or movement of the steel rod, depending on the angle and other factors such as the material and weight distribution of the rod.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
822
  • Introductory Physics Homework Help
Replies
8
Views
6K
  • Introductory Physics Homework Help
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
12
Views
5K
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
22
Views
2K
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Back
Top