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Homework Help: Magnetic force in wire

  1. Mar 25, 2006 #1
    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?
  2. jcsd
  3. Mar 26, 2006 #2
    Can someone please help me?
  4. Mar 26, 2006 #3


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    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".
  5. Mar 27, 2006 #4
    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...
  6. Mar 27, 2006 #5


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    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)

  7. Mar 27, 2006 #6
    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
  8. Mar 27, 2006 #7


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    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: Mar 27, 2006
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