What is the Magnitude of the Force on a Wire in a Uniform Magnetic Field?

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Homework Help Overview

The discussion revolves around calculating the magnitude of the force on a wire segment carrying a current in a uniform magnetic field. The problem involves vector quantities, including the current, the length of the wire, and the magnetic field strength, which are essential for determining the force using the cross product.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the cross product in calculating the force, with some confusion regarding the correct use of vector multiplication versus scalar multiplication. There are attempts to clarify the calculation of the cross product and its implications for the force's direction and magnitude.

Discussion Status

Several participants are actively engaging in the calculations, with some providing corrections to earlier attempts. There is a focus on understanding the vector nature of the quantities involved, and while some guidance has been offered regarding the cross product, there is still uncertainty about the calculations and results.

Contextual Notes

Participants are addressing potential misunderstandings about the values used in calculations, particularly the magnetic field strength. There is also a discussion about the treatment of the results from the cross product, indicating a need for clarity on vector operations.

Punchlinegirl
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A current of 3.1 A flows in a straight wire segment length (4 cm)i + (5 cm)j in a uniform magnetic field of strength (0.9 T) i. Find the magnitude of the force on the wire. Answer in units of N.
First I found the magnitude of the length of the wire, [tex]\sqrt .04^2 +.05^2[/tex] = .0640 m.
Then I used the equation F= IL x B
(3.1)(.0640)(.09)= .0179 N
This isn't right.. I don't really know how to do these problems when both the field and length have directions.. can someone please help?
 
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Can someone please help me?
 
The x in F=ILxB is a cross product, not a simple multiplication. It multiplies vectors, not numbers. Look up how to do a "cross product".
 
Ok so when I do the cross product, I get -.01395. Then if I multiply it by the current, 3.1 A, so I get .01395, since they want the magnitude.. which still isnt' right...
 
Punchlinegirl said:
Ok so when I do the cross product, I get -.01395. Then if I multiply it by the current, 3.1 A, so I get .01395, since they want the magnitude.. which still isnt' right...

Can you show how you do a cross product? Your answer for the cross product is incorrect (and I am not sure how you got 3.1 times .01395 =0.01395)

Pat
 
I did
i j
.04 .05
.09 0
so .04*0 = 0
.09 * .05 = .0045
0-.0045= -.0045

Then -.0045 * 3.1 = .01395, since it's absolute value
 
Punchlinegirl said:
I did
i j
.04 .05
.09 0
so .04*0 = 0
.09 * .05 = .0045
0-.0045= -.0045

Then -.0045 * 3.1 = .01395, since it's absolute value

First, why do you use 0.09 for B? It's 0.9.
Also, it seems like you treat th eresult of the cross product as if it was a bunch of numbers to be added together. You cannot do that, a cross product gives a vector!

[itex]{\vec L} \times {\vec B} = (.04m {\vec i} + 0.05m {\vec j}) \times (0.9 T{\vec i}) = -0.045 T \cdot m {\vec k}[/itex]
So the force will be [itex]-0.1395 N {\vec k}[/itex]

Edit: and of course the *magnitude* of the force will be 0.1395 N. My point was that the difefrent terms obtained from the cross product can't simply be added together.
 
Last edited:

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