# Magnetic Force on Various Sides of Spinning Loop?

Magnetic Force on Various Sides of Spinning Loop??

## Homework Statement

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?

## Homework Equations

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

## 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!

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collinsmark
Homework Helper
Gold Member

I'm using Fb=ILBSsintheta, and found theta to be 25 degrees.
$$\vec{F_B} = \vec IL \times \vec B,$$
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.)
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
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.