# 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