Find the average force on the loop
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Find the average force on the loop

[From: ] [author: ] [Date: 11-11-24] [Hit: ]
centred on (0,0,h) and parallel to the xy plane.The distance from the proton is a.Electrical potential energy(PE) = e(-e)/(4πεa) =- e²/(4πεa)Total energy E = PE + KEKE = E - PE ½mv²= E - (-e²/(4πεa))v² = (2/m)(E + e²/(4πεa))v = √ [2/m)(E + e²/(4πεa))]Since x² + y² = a and z = h√ (x² + y² + z² ) = √(a² + h²)So the expression for is = [B/√(a² + h²)] [x + y +h]The system is symmetrical about the z-axis.That means,......
http://www.cramster.com/answers-nov-11/physics/average-force-loop_1708675.aspx?rec=0

where does the term a come into effect. How does the cross product yield a. I'm very close to the answer, but I'm missing the a.


And another question.

When viewed from the origin, is the electron moving clockwise or counterclockwise, and why?

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I’ll denote vectors using angle brackets and use ‘a’ rather than ‘a0’ and ‘ε’ rather than ‘ε0’, for neatness.

From the data supplied you can tell that the electron is moving in a circle radius a, centred on (0,0,h) and parallel to the xy plane. The distance from the proton is a.

Electrical potential energy (PE) = e(-e)/(4πεa) = - e²/(4πεa)

Total energy E = PE + KE
KE = E - PE
½mv²= E - (-e²/(4πεa))
v² = (2/m)(E + e²/(4πεa))
v = √ [2/m)(E + e²/(4πεa))]

Since x² + y² = a and z = h
√ (x² + y² + z² ) = √(a² + h²)
So the expression for is
= [B/√(a² + h²)] [x + y +h]

The system is symmetrical about the z-axis. That means, over one orbit, the average resultant force in the x and y directions will be zero; this is the force produced by the z-component of B; so the z-component of B has no effect on the average force.
We therefore need only consider the effects of the x and y components s of B.

At any point in the loop, when the electron is at coordinates (X, Y, h,) the magnetic field is
= [B/√(a² + h²)] [X + Y +h]
Since we can ignore the z-xomponent, this means the relevant force is
: = [B/√(a² + h²)] [X + Y]
and this has magnitude
|B| = {B/√(a² + h²)] √ [X²+ Y²]
= {B/√(a² + h²) √ [a²]
= aB/√(a² + h²)
(This is where the ‘a’ comes from.)

This is a constant magnitude force. And because of symmetry (no orientation is preferred) it must act radially. In fact it acts radially outwards since it is of the form x + y rather than -x -y.
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