Magnetic Force on Electric Currents

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A wire carrying a current experiences a magnetic force when placed in a magnetic field, as described by the equation F=ILB. In this case, with a force of 0.810 Newtons, a wire length of 2.00 meters, and a magnetic field strength of 0.030 Teslas, the current can be calculated. The correct formula application shows that the current I should be derived from I=F/(LB), leading to I=0.810/(2.00*0.030), which equals 13.5 A, assuming the wire is oriented at a 90-degree angle to the magnetic field. There was confusion in the initial calculations, particularly in the algebraic manipulation, which could lead to loss of marks in an academic setting. Proper orientation of the wire relative to the magnetic field is crucial for accurate results.
Mikala
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1. A current in a wire which has a length of 2.00m in a magnetic Field of 0.030 Teslas, feels a magnetic force of 0.810 Newtons. What is the current in the wire?2. F=IlB, Find I3. F= 0.810, l= 2.00m, B= 0.030 T, I=? (A)
I= l x B I= 2.00m x 0.030 T I= 0.074 A
F 0.810
Not really sure if the setup is correct.
 
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The question is poorly worded ... it should be more that 2m of a wire carrying a current passes through a 0.030T uniform magnetic field and experiences a force of 0.810N... what is the current? (Missing: orientation of the wire to the field.)

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html
You appear to have started with the correct approach: F=ILB
You know F, L, and B: so solve for I and plug the numbers in.
(assuming the current is oriented 90deg to the field.)

I don't know why you wrote I=lxBI or how you got I=0.074A
Because of the way you have written your algebra, your thinking is unclear: this would cost you marks.
 
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