# Homework Help: Magnetic Force on Various Sides of Spinning Loop?

1. Mar 29, 2012

### longcatislong

Magnetic Force on Various Sides of Spinning Loop??

1. The problem statement, all variables and given/known data

A horizontal uniform magnetic field of magnitude 0.4 T is oriented at an angle of = 65° relative to a line perpendicular to plane of a vertical, rectangular loop, as indicated in the figure. The loop has a height of 8 cm and a width of 15 cm. Each of the 1000 turns of wire in the loop carries a current of 0.22 A counterclockwise around it.

a link to the diagram: http://www.webassign.net/userimages/last-prob-diag-small.jpg?db=v4net&id=86188

(a) Find the size of the force (+ only) on each side of the loop:
top: ____N?
bottom: 2 N
left: ___N?
right:____N?

2. Relevant equations

Fb=ILBsintheta (where theta is the angle between I and B)

3. The attempt at a solution

I'm using Fb=ILBSsintheta, and found theta to be 25 degrees. For the force on the top side, i'm using the length of the top side, 15cm=.15m.

I=.22A
L=.15m
B=.4T
theta=25 degrees

So, I get Fb=(.22)(.25)(.4)(Sin25)=.005579

I've also tried using the length of the other side, but it's still wrong according to my online homework.

Anything will help! This assignment is due soon!

2. Mar 29, 2012

### collinsmark

Re: Magnetic Force on Various Sides of Spinning Loop??

$$\vec{F_B} = \vec IL \times \vec B,$$
which can be represented by
$$F_B = ILB \sin \theta,$$
where $\theta$ is the angle between $\vec I$ and $\vec B$. Here I is the total current on a given side, whatever the configuration of wires carrying the total current happens to be.

So for some of sides of the loop, $\theta$ is 25 degrees. But not all sides! (Take it one side at a time. I caution against using 25 deg all willy-nilly.)
The I that you are using (0.22 A) is for a single turn of wire. But there are 1000 turns of wire in the loop. So you're missing a factor of 1000 in there somewhere.