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

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SUMMARY

The magnitude of the force on a wire carrying a current of 3.1 A in a uniform magnetic field of strength 0.9 T can be calculated using the formula F = IL x B. The wire segment has a length of 0.0640 m, derived from its components (4 cm, 5 cm). The correct application of the cross product yields a force of 0.1395 N. Misinterpretations of vector operations and incorrect values for magnetic field strength led to initial calculation errors.

PREREQUISITES
  • Understanding of vector mathematics, specifically cross products
  • Familiarity with magnetic fields and forces on current-carrying conductors
  • Knowledge of SI units, particularly for current (A), magnetic field (T), and force (N)
  • Basic proficiency in physics equations related to electromagnetism
NEXT STEPS
  • Study vector cross product calculations in detail
  • Learn about the Lorentz force law and its applications
  • Explore the implications of magnetic fields on current-carrying wires
  • Investigate the effects of varying magnetic field strengths on force magnitude
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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, \sqrt .04^2 +.05^2 = .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!

{\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}
So the force will be -0.1395 N {\vec k}

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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