Magnetic force in wire

1. Mar 25, 2006

Punchlinegirl

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, $$\sqrt .04^2 +.05^2$$ = .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. Mar 26, 2006

Punchlinegirl

3. Mar 26, 2006

krab

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".

4. Mar 27, 2006

Punchlinegirl

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...

5. Mar 27, 2006

nrqed

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)

Pat

6. Mar 27, 2006

Punchlinegirl

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

7. Mar 27, 2006

nrqed

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!

${\vec L} \times {\vec B} = (.04m {\vec i} + 0.05m {\vec j}) \times (0.9 T{\vec i}) = -0.045 T \cdot m {\vec k}$
So the force will be $-0.1395 N {\vec k}$

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