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Old Apr 18th 2018, 01:46 PM   #1
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What is the magnitude of the magnetic field?



ok this is my overleaf file

but not sure how to solve this

mahalo much
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Old Apr 19th 2018, 04:36 AM   #2
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The force on a proton, which has charge 1, traveling with velocity vector v through a magnetic field with vector u is the cross product of u and v. That is mdv/dt= v x u.

If the magnetic field vector is $\displaystyle \left<M_x, M_y, M_z\right>$ and the proton's velocity vector is $\displaystyle \left<v_x, v_y, v_z\right>$ then
$\displaystyle \left<m\frac{dv_x}{dt}, m\frac{dv_y}{dt}, m\frac{dv_z}{dt}\right>= \left<M_yv_z- M_zv_y, M_zv_x- M_xv_z, M_xv_y- M_yv_x\right>$.

So you need to solve the three equations
$\displaystyle m\frac{dv_x}{dt}= M_Yv_z- M_zv_y$
$\displaystyle m\frac{dv_y}{dt}= M_zv_x- M_xv_z$
$\displaystyle m\frac{dv_z}{dt}= M_xv_y- M_yv_x$

If you set up a coordinate system so that the magnetic field vectors are in the y direction $\displaystyle \left<M_x, M_y, M_z\right>= \left<0, M_y, 0\right>$ so those equations become
$\displaystyle m\frac{dv_x}{dt}= M_yv_z$
$\displaystyle m\frac{dv_y}{dt}= 0$
$\displaystyle m\frac{dv_z}{dt}= -M_yv_x$
quite a lot simpler!

For example, if you differentiate the first equation again,
$\displaystyle m\frac{d^2v_x}{dt}= M_y\frac{dv_z}{dt}$
and you can replace that derivative of $\displaystyle \frac{dv_z}{dt}$ from the third equation to get $\displaystyle m^2\frac{d^2v_x}{dt^2}= -M_y^2v_x$
a relatively easy second order linear differential equation with constant coefficients- though certainly not secondary school level!

Probably you are given a formula relating the velocity and magnetic field strength of a "radius of curvature". Do you have such a formula?

Last edited by HallsofIvy; Apr 19th 2018 at 04:39 AM.
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Old Apr 20th 2018, 05:11 AM   #3
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Originally Posted by countdown View Post


ok this is my overleaf file

but not sure how to solve this

mahalo much
The magnetic force on a charged particle where the velocity is perpendicular to the field, is

$\displaystyle F = Bqv$

Then, because the particle is undergoing circular motion, you have

$\displaystyle F = \frac{mv^2}{r}$

Equating them should allow you to get an equation for the magnetic flux density in terms of known parameters of the charged particle and the variables describing the motion of the particle.

FYI: this is cyclotron motion
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Old Apr 22nd 2018, 12:14 AM   #4
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thanks everyone
that was great help
glad i came here
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