How Does Small Angle Approximation Affect Magnetic Field Calculations?

[Firben]

Homework Statement

I want to solve the motion equation

$m \frac {dv_z} {dt} = - μ \frac {∂B_z} {∂z}$
with small angle approximation

Homework Equations

$B_z(z) = B_0 -bCos(\frac {zπ} {2L})$ is the magnetic field in the z-direction

The Attempt at a Solution

Started by derive the magnetic field equation with resp. z
$m \frac {∂B_z} {∂z} = \frac {bπ} {2L} * μ * sin(\frac {zπ} {2L})$
and put it into eq above <=>

$m \frac {dv_z} {dt} = -μ \frac {bπ} {2L} * sin(\frac {zπ} {2L})$
<=>
small angles sin(x) ~ x
$\frac {dv_z} {dt} = - \frac μ m * \frac {bπ} {2L} * z$

$v_z = \frac {dz} {dt}$ <=>

$\frac {d^2v_z} { dt^2} = - \frac {μb} {m} * \frac {μ} {2L} * z$

Im not sure if this is correct

 

[Charles Link]

Homework Statement

I want to solve the motion equation

$m \frac {dv_z} {dt} = - μ \frac {∂B_z} {∂z}$
with small angle approximation

Homework Equations

$B_z(z) = B_0 -bCos(\frac {zπ} {2L})$ is the magnetic field in the z-direction

The Attempt at a Solution

Started by derive the magnetic field equation with resp. z
$m \frac {∂B_z} {∂z} = \frac {bπ} {2L} * μ * sin(\frac {zπ} {2L})$
and put it into eq above <=>

$m \frac {dv_z} {dt} = -μ \frac {bπ} {2L} * sin(\frac {zπ} {2L})$
<=>
small angles sin(x) ~ x
$\frac {dv_z} {dt} = - \frac μ m * \frac {bπ} {2L} * z$

$v_z = \frac {dz} {dt}$ <=>

$\frac {d^2v_z} { dt^2} = - \frac {μb} {m} * \frac {μ} {2L} * z$

Im not sure if this is correct

You dropped a factor of $2 L$ in the denominator, as well as $\pi$ in the numerator, when you took the small angle approximation. (Also, your first equation should read $\mu \frac{\partial{B}}{\partial{z}}$, (instead of $m$), but that is clearly a "typo"). Your last equation needs to read $\frac{d^2 z}{dt^2}$, and please check your algebra on this. I believe you left off the $\pi$,(which should actually be $\pi^2$), and you have an extra $\mu$. $\\$ Note: $\sin(\frac{z \pi}{2 L})=\frac{z \pi}{2 L }$ for small angles. $\\$ Editing: Did they give you any initial conditions?=initial position and velocity? Otherwise, I guess you assume $z=0$ and $v_z=0$ at time $t=0$. These initial conditions, though, would lead to the trivial solution $z=0$ for all $t$ if I'm not mistaken.

 

[Ray Vickson]

Homework Statement

I want to solve the motion equation

$m \frac {dv_z} {dt} = - μ \frac {∂B_z} {∂z}$
with small angle approximation

Homework Equations

$B_z(z) = B_0 -bCos(\frac {zπ} {2L})$ is the magnetic field in the z-direction

The Attempt at a Solution

Started by derive the magnetic field equation with resp. z
$m \frac {∂B_z} {∂z} = \frac {bπ} {2L} * μ * sin(\frac {zπ} {2L})$
and put it into eq above <=>

$m \frac {dv_z} {dt} = -μ \frac {bπ} {2L} * sin(\frac {zπ} {2L})$
<=>
small angles sin(x) ~ x
$\frac {dv_z} {dt} = - \frac μ m * \frac {bπ} {2L} * z$

$v_z = \frac {dz} {dt}$ <=>

$\frac {d^2v_z} { dt^2} = - \frac {μb} {m} * \frac {μ} {2L} * z$

Im not sure if this is correct

No, it is not. You dropped some factors, and your DE should be for $d^2 z/dt^2$, not $d^2 v_z / dt^2$.