# Calculating the magnetic force on a charged particle

For my homework I'm supposed to answer the following:

Now consider the example of a positive charge q moving in the xy plane with velocity v⃗ =vcos(θ)i^+vsin(θ)j^ (i.e., with magnitude v at angle θ with respect to the x axis). If the local magnetic field is in the +z direction, what is the direction of the magnetic force acting on the particle?

Now, I know that the answer is the determinate of the matrix \begin{pmatrix} i & j & k \\ vcos(\theta)i & vsin(\theta)j & 0 \\ 0 & 0 & k \end{pmatrix}

but for the life of me I cannot figure the details of calculating the value. Specifically, I don't know what happens when I multiply the unit vectors and how and when they should appear in the matrix. How do you do this? Also, is this the right matrix?

If you're set on using the matrix-determinant method (which isn't any sort of problematic, but isn't the only way to do this), then your matrix should read:

$\vec{F}_B=q\left|\matrix{\hat{x}&\hat{y}&\hat{z}\\v\cos\theta&v\sin\theta&0\\0&0&B}\right|$

This would give you:

$\vec{F}_B=q\left(\vphantom{\dfrac{a}{b}}\hat{x}\left[vB\sin\theta\right]+\hat{y}\left[-vB\cos\theta\right]\right)$

To find the direction only, you can divide $\vec{F}_B$ by its own magnitude to give you a unit vector that points in the direction of the magnetic force:

Direction $=\hat{F}_B=\dfrac{\vec{F}_B}{\left|\vec{F}_B\right|}$

When you use a matrix for cross products, you don't actually put the i's, j's and k's in the matrix, except for the first line. So the matrix you want to take the determinant of is the same except remove the unit vectors in the second and third rows. Remember the whole matrix thing is simply a mnemonic and not really mathematically rigourous.

• so should the last row be (0,0,1)? – Zachary F Feb 18 '16 at 17:58
• Yes, that is correct – bremsstrahlung Feb 18 '16 at 17:59
• Okay. But then I still end up with $(vsin(\theta)i-vcos(\theta)j)k$. Do I just multiply them in the order they appear? – Zachary F Feb 18 '16 at 18:01
• Your $k$ is wrong. And the result is a vector. Vectors have direction. Compare the direction of $\vec{v}$ with the direction of $\vec{F}$. – Bill N Feb 18 '16 at 18:12

I think in the previous answer the focus has been too much on the method of using a determinant to calculate the cross product. But the question seems to be asking why one needs to calculate the given matrix determinant.

You are given a magnetic field which is assumed constant and homogenuous, so we write down $$\overrightarrow{B}(\overrightarrow{x},t) = B_0 \overrightarrow{e}_z,$$ where $\overrightarrow{e}_z$ is the unit vector in $z$-direction. Your charged particle has a velocity given by $$\overrightarrow{v} = v_0 \cos(\theta) \overrightarrow{e}_x + v_0 \sin(\theta) \overrightarrow{e}_y.$$ Now the "magnetic force" pulling on the charged particle is known as the Lorentz force and given in general by $$\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{v}\times\overrightarrow{B}),$$ where $q$ is the charge and $\overrightarrow{v}$ is the velocity of the particle in the electromagnetic field. In your situation, you only have a magnetic field present. So for the Lorentz force you need to calculate the cross product of $\overrightarrow{v}$ with $\overrightarrow{B}$. This will give you the direction and magnitude of the Lorentz force, as asked in your question. Which method you choose to calculate the cross product, for example using the mentioned determinant method, is up to you. If you're asking why the mentioned determinant gives the wanted crossproduct, I guess you find the answer with little research effort e.g. on wikipedia.

the method of using matrix, is just to calculate the cross product of two vectors in given components along 3 perpendicular axes, what you need to understand is just the vector products of two perpendicular unit vectors

this image nails it on the head, imagine the x, y and z axis and proceed in the above way until it becomes like the back of your hand,

(source: millersville.edu)

when you multiplay i and j , you are actually doing this ->

(source: kshitij-iitjee.com)