Finding the magnitude of the current flowing through a wire

[rocky4920]

Homework Statement

Homework Equations

The Attempt at a Solution

[/B]

My question: in order to find the answer for part d, I have the L, the B, and the mass. However, the F equations have I, which I do not have. Would I have to solve for I in order to find F? If so, I would like some guidance on which formula is best to tackle part d.
Thank you for your time.

 

[gneill]

Part (d) is asking you to find the current...

 

[rocky4920]

So by doing so I would manipulate the following formula: B = µo I/2 pi r ?

I = 2 pi r B /

µo​

r = 0.25m since wire is .50 m
B= 1.6 T

µo​

= 4 pi x 10-7​

 

[gneill]

Um, no. The formula you've chosen gives the strength of the magnetic field at some distance r from a long current-carrying wire. And it doesn't make sense to take the distance to be the length of a short wire segment.

Instead, look at the reason for your answer to part (b).

 

[rude man]

Your drawing of the B field is incorrect. You've drawn the B field generated by the wire but that won't hold up the wire!

 

[rocky4920]

Your drawing of the B field is incorrect. You've drawn the B field generated by the wire but that won't hold up the wire!

Based on the right hand rule, though, wouldn't the field go out of the page above the wire, and then into the page below it?

 

[rocky4920]

Um, no. The formula you've chosen gives the strength of the magnetic field at some distance r from a long current-carrying wire. And it doesn't make sense to take the distance to be the length of a short wire segment.

Instead, look at the reason for your answer to part (b).

Thank you for the help. Yes, I definitely was using the wrong equation.

F= ILA, therefore, I= F/LB
Is that a better approach to use?

 

[gneill]

Thank you for the help. Yes, I definitely was using the wrong equation.

F= ILA, therefore, I= F/LB
Is that a better approach to use?

Yes, but... Why is the wire in equilibrium? What forces are balanced?

 

[gneill]

Based on the right hand rule, though, wouldn't the field go out of the page above the wire, and then into the page below it?

As stated by @rude man , the wire won't hold itself up, regardless of the field it itself produces. What other field is mentioned in the problem statement? How must it be oriented in order to produce the required lift?

 

[rocky4920]

Yes, but... Why is the wire in equilibrium? What forces are balanced?

FE and FB. FE-FB= zero.
The electric field and the magnetic field.

 

[gneill]

FE and FB. FE-FB= zero.
The electric field and the magnetic field.

No, there is only magnetic field and mass mentioned in the problem statement.

 

[rocky4920]

 

[gneill]

The link to the image that you've posted (post #12) is broken or is on a site that does not allow public access. PLease upload a copy of your image to our servers (UPLOAD button).

 

[rocky4920]

Photo: