What is the force experienced by a wire in a magnetic field?

[jimmyboykun]

Homework Statement

A long wire carrying 6.50A of current makes two bends, as shown in the figure (Figure 1) . The bent part of the wire passes through a uniform 0.280T magnetic field directed as shown in the figure and confined to a limited region of space.

http://session.masteringphysics.com/myct/itemView?assignmentProblemID=31713281&offset=next

Part A: Find the magnitude of the force that the magnetic field exerts on the wire.
Part B: Find the direction of the force that the magnetic field exerts on the wire.

Homework Equations

F=IlB

The Attempt at a Solution

for part A I calculated the force by the given information.
I= 6.50A, B= 0.280T, and l=0.8m(I did the Pythagorean theorem to find the lenght)

F=(6.50A)(0.8m)(0.280T)

F= 1.456N.

I found the force, but I got the answer wrong. Can someone explain where did I go wrong? I figured that I have the length incorrect, but I don't know how to find the length.

 

[Larry Gopnik]

Hi,

Is it possible to see a screenshot of the question with the diagram? The link doesn't work.

 

[Simon Bridge]

Yep - link does not help: "not signed in" message.

 

[jimmyboykun]

I'm sorry here's another link http://s16.postimg.org/li2tzpa4l/yg_20_92.jpg

 

[Simon Bridge]

OK - so you want $\vec F = q\vec v \times \vec B = L\vec I \times \vec B$
Notice the cross product there?
See: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/forwir2.html

But from what I can see of your working - it looks like you left off part of the wire.

Note: the diagonal section makes a 1-2-√3 triangle,
so sin(30)=cos(60)=1/2, cos(30)=sin(60)=(√3)/2

The opposite side is O=40cm by trigonometry: O = Hsin(30) = H/2 means H = 80cm.

It's a good idea to look out for these special triangles.