Is the Right-Hand Rule Applied Correctly in Determining Magnetic Forces?

[LaneRendell]

Homework Statement

Homework Equations

The Right-Hand Rule

The Attempt at a Solution

I'm having some issues with the right-hand rule to find the direction of magnetic force, and I'm doing a homework problem.
The answers I got are (letters are vector directions):

A: -j
B: i
C: j
D: i
E:-i
F:-i

Am I doing this right? If not could someone explain a good way to do the right hand rule or where I'm making the mistake. As I understand it, you point your uncurled palm in the direction of the current and then curl your fingers in the direction of the magnetic field.

 

[Psinter]

Hi,

Actually, all your fingers except the thumb must point in the direction of the current and then you curl them in the direction of the magnetic field. The thumb will indicate the direction of the force once your fingers are curled. If you are already pointing in the direction of the magnetic field and have no need to curl your fingers, then the force and direction are 0. In segment C, your fingers point to the magnetic field direction (z), therefore you got that one wrong, the answer is $\vec{0}$. Verify segment D and segment F. The rest are okay.

I can help you to verify algebraically. Maybe someone with more experience can help with the right hand rule since I too have a little trouble with it. Algebraically we can check that at segments: C, D, and F; you got it wrong. The rest are correct (A, B, E). Here is an explanation, but you can skip to the verification examples I wrote if you find the explanation too verbose and learn better by looking at examples.

We know that: $\vec{F} = q\vec{v} \times \vec{B}$

For the purpose of this exercise we ignore q.
We will take $\vec{v}$ to mean the direction vector of our current. Since we are working in a unit cube we will always assume the magnitude of $\vec{v}$ to be 1.
$\vec{B} = (0i + 0j + 1k)$, the direction vector of our magnetic field in this specific exercise.

Now before I give you examples of verification, we note that since we have a cross product of the direction of the current and the direction of the magnetic field, any current moving in the same direction as the magnetic field will experience 0 force. Or mathematically speaking: $\left | \vec{F}\right | = q\left |\vec{v}\right | \left | sin(\theta) \right | \left | \vec{B}\right |$

If you look at that, when the current moves in the same direction as the magnetic field, $\theta = 0$, which makes the whole equation 0.

Verification Examples:
I will make examples of verification for A, B, and C so you can get the hang of it:

$\vec{F_{A}} = (1i + 0j + 0k) \times (0i + 0j + 1k)$

$\begin{vmatrix} i & j & k \\ 1 & 0 & 0 \\ 0 & 0 &1 \end{vmatrix} = \begin{vmatrix} 0 & 0 \\ 0 &1 \end{vmatrix}i - \begin{vmatrix} 1 & 0\\ 0 &1 \end{vmatrix}j + \begin{vmatrix} 1 & 0\\ 0 &0 \end{vmatrix}k$

$= (0*1 - 0*0)i - (1*1 - 0*0)j + (1*0 - 0*0)k$
$= 0i - 1j + 0k$
$\vec{F_{A}} = -j$

$\vec{F_{B}} = (0i + 1j - 1k) \times (0i + 0j + 1k)$

$\begin{vmatrix} i & j & k \\ 0 & 1 & -1 \\ 0 & 0 &1 \end{vmatrix} = \begin{vmatrix} 1 & -1 \\ 0 &1 \end{vmatrix}i - \begin{vmatrix} 0 & -1\\ 0 &1 \end{vmatrix}j + \begin{vmatrix} 0 & 1\\ 0 &0 \end{vmatrix}k$

$=(1 * 1 - -1*0)i - (0*1 - -1*0)j + (0*0 - 1*0)k$
$= 1i - 0j + 0k$
$\vec{F_{B}} = i$

$\vec{F_{C}} = (0i + 0j + 1k) \times (0i + 0j + 1k)$

$\begin{vmatrix} i & j & k \\ 0 & 0 & 1 \\ 0 & 0 &1 \end{vmatrix} = \begin{vmatrix} 0 & 1 \\ 0 &1 \end{vmatrix}i - \begin{vmatrix} 0 & 1\\ 0 &1 \end{vmatrix}j + \begin{vmatrix} 0 & 0\\ 0 &0 \end{vmatrix}k$

$=(0*1 -1*0)i - (0*1 -1*0)j + (0*0 - 0*0)k$
$= 0i -0j + 0k$
$\vec{F_{C}} = \vec{0}$

As you will expect, $\vec{F_{C}} = \vec{F_{F}} = \vec{0}$, because both of currents in those segments move in the same direction as the magnetic field.