00:01
For part a, we're asked to find an expression for the magnitude of the magnetic field and determine the direction of the magnetic field.
00:09
Okay, so the current through each turn is going to be i.
00:13
There are in turns in the coil, therefore the total current in the coil is n times i.
00:18
This current travels towards the right in the lower side of the coil, and the magnetic force acting on the lower side of the coil, which we call f sub m.
00:28
Let's restart that.
00:32
Is going to be equal to the magnitude of the magnetic field, b, times the number of turns in the coil, times the current in the coil, times the width, w, times the sign of the angle, theta.
00:48
So again, b is the strength of the magnetic field, w is the width of the coil, and theta is the angle between the magnetic field vector and the direction of the current.
00:54
But since the field is perpendicular to the direction of the current, the angle theta, is going to be 90 degrees.
00:59
So if theta's 90 degrees, then this is just equal to b times n, times i times w so the magnetic field must be directed out of the page for the magnetic force to be downed so we can say in order for the magnetic force to be down the b field must be directed out of the page so part of our solution for a was to determine the directionality so out of the page is part of that solution so we can box it in but we're also asked to find an expression for the the magnitude of the magnetic field.
01:43
So we still need to work on that part.
01:45
Well, the gravitational force acting on the coil is going to be an equilibrium with the magnetic force.
01:50
So they're going to be balanced.
01:52
So we can say that f sub m in this case is equal in magnitude to f sub g, or f sub g is the force of gravity.
02:01
So we can just replace these expressions with what we know for the force of the magnetic field, which we know to be b times n times i times w, and the force of gravity is just mass times the acceleration of gravity...