Electromagnetism Question

1. "A wire segment carries a current of 1.8A from a to b. There is a uniform magnetic field of 1.2T acting in the z-direction. Find the total force on the wire and show that the total force is the same as if the wire were a straight segment from a to b".

I've attached the diagram showing what the segment looks like and I've called the point between a and b, c.

2. Relevant equations: F=BIL

3. Would this be right?:
Fac = ILBcos90 = 0N (90° since the x-direction is orthogonal to z)
Fcb = ILBsin90 = ILB = 1.8 x 0.04 x 1.2 = 0.0864N (90° since the y-direction is orthogonal to z)
=> Facb = 0.0864N

Fab = ILBsinθ = ILBsin(arctan(4/3)) = 0.8ILB = 0.8 x 1.8 x 0.05 x 1.2 = 0.0864N

Therefore, Facb = Fab

Or is it just a coincidence that the two values come out to be the same? Because I don't really understand why it would be cos90 from a to c and sin90 from c to b =s



Staff Emeritus
Science Advisor
Homework Helper
Welcome to PF :smile:

There are some problems with your derivation. For one, all currents (ac, cb, and ab) are perpendicular to the magnetic field. So use F=BIL to find each force, with no cos or sin factors. None of the three forces will be zero.

You'll also need to use the right-hand rule to figure out the direction of each force, and combine the ac & cb forces.
Thanks for replying,
So, would this be right:
Fac = ILB = 0.0648N in the -ve y-direction so -0.0648j
Fcb = ILB = 0.0864N in the +ve x-direction so 0.0864i
=> Facb = 0.0864i - 0.0648j

Fab = ILB = 0.108N
The direction is sin(arctan(4/3))Fi - cos(arctan(4/3))Fj
= 0.0864i - 0.0648j

Not sure about the signs though because the right-hand rule really confuses me =s
Last edited:


Staff Emeritus
Science Advisor
Homework Helper
Looks good! The signs are correct too.
Oh good :smile:
Thanks a lot.

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