anonymous
  • anonymous
A 10m wire that carries a current of 3A from east to west is placed in a magnetic field of 0.0004T running from south to north. Find the size and direction of the magnetic force on the wire. **how do we find this? :/ Thanks!!:)
Physics
schrodinger
  • schrodinger
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Michele_Laino
  • Michele_Laino
the situation described into your problem is: |dw:1433612353948:dw|
anonymous
  • anonymous
ok!
Michele_Laino
  • Michele_Laino
since magnetic field and current are perpendicular each other, then the magnitude of the magnetic force is: \[\Large F = ILB\] where L = 10 meters

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anonymous
  • anonymous
yes:)
Michele_Laino
  • Michele_Laino
namely: \[\Large F = ILB = 3 \times 10 \times 4 \times {10^{ - 4}}\]
anonymous
  • anonymous
ok so we get 0.012?
Michele_Laino
  • Michele_Laino
yes! correct the unit of measure is Newtons, so we have: F= 0.012 Newtons
anonymous
  • anonymous
is that the size of the force? :/
Michele_Laino
  • Michele_Laino
yes! now the direction is given by the rule of product vector, so the orientation of the magnetic force is like below: |dw:1433612811550:dw| namely the vector F is perpendicular to both current and magnetic field, and it is oriented inward of the drawing plane
anonymous
  • anonymous
ok! so it is going south?
Michele_Laino
  • Michele_Laino
no, it is going inward of the drawing plane, here is a 3-D view: |dw:1433613109921:dw|
anonymous
  • anonymous
ohh so it is going northeast?
Michele_Laino
  • Michele_Laino
no, that is a 3-D view not a 2-D view
anonymous
  • anonymous
ohh i am confused then haha what direction is that ? :/
Michele_Laino
  • Michele_Laino
it is perpendicular with respect to North-South direction and it also is perpendicular with respect to West-East direction
anonymous
  • anonymous
ohh okay :/ wow confusing hahah :P thank you!! so this problem is complete?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
yay thank you1!:D
Michele_Laino
  • Michele_Laino
:)

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