# An electron moving through a uniform magnetic field

1. Apr 7, 2015

### RyanTAsher

1. The problem statement, all variables and given/known data

An electron moves through a uniform magnetic field given by B = Bx i + (3.68 Bx) j . At a particular instant, the electron has velocity v = (1.88 i +4.86 j) m/s and the magnetic force acting on it is (2.43 × 10-19) N. Find Bx.

Givens

B = [Bx i + (3.68 Bx)]

v = (1.88 + 4.86)

Fb = (2.43 × 10-19) N. Find Bx.

2. Relevant equations

Fb = q(B x V)

3. The attempt at a solution

(2.43 x 10^-19)k = (1.602 x 10^-19) [(Bx i + (3.68Bx) j) x (1.88 i + 4.86 j)

cross product resultant:

(2.43 x 10^-19)k = (1.602 x 10^-19) (-2.06*Bx)k

(2.43 x 10^-19)k = (-3.30 x 10^-19*Bx) k

I don't understand how to solve for the Bx, we haven't learned how to divide vectors, and I can't really pull the Bx out of the k vector as a scalar. How do I solve for the Bx???

2. Apr 7, 2015

### haruspex

Since the two unit vectors are identical, you can cancel them. But if you want to do it by pure methods, there are two ways: consider magnitudes; take the dot product of each side with something convenient.

3. Apr 7, 2015

### BvU

BX is not a vector, it is a number with a dimension. The vectors are $\vec v$ and $\vec B = B_X \hat\imath + (3.68\;B_X) \hat\jmath\$.
So check the dimensions (they should be fine) and find the value as you intended to do: by dividing.

4. Apr 7, 2015

### haruspex

I believe Ryan is referring to the two k vectors.

5. Apr 7, 2015

### collinsmark

Hello RyanTAsher,

Before I go through the math (privately on my own), could you help clear up a few ambiguities?

I'm assuming that the x in Bx is a subscript and not a variable. (In other words, I'm assuming that $\vec B$ is uniform and not a function of x.) Is that correct?

Are you sure about $\vec F = q \left( \vec B \times \vec v \right)$? Is that correct or is something backwards?

(Also, don't forget that the charge of an electron is negative. i.e., $q = -e$)

In the initial problem statement, the force was given as (2.43 × 10-19) N with no direction specified. Later, you tacked on a $\hat k$. I'm just wondering if the original problem statement had the $\hat k$ in it, or if the (2.43 × 10-19) N figure was meant to be magnitude only without specifying a direction. [Edit: Going through the math, I realize that the force must be along the z axis. But ultimately, one still needs to determine if the force is pointing in the positive or negative direction of the axis.]

Last edited: Apr 7, 2015
6. Apr 7, 2015

### BvU

@Haru: of course, naive of me to overlook that; thanks!
@CM: well pointed out. But now the wrong expression is appearing twice already in this thread (shudder) ....

@RTA: you're in good hands ! Bedtime for me. Good luck. And dividing out vectors is like dividing out Yuzzamatuzzes: if Zfluff Yuzzamatuzzes = 2.5 Yuzzamatuzzes, then you can be sure Zfluff = 2.5. It's a property of multiplication, not a property of Yuzzamatuzzes. There is only one snag to beware of: things go awry if Yuzzamatuzz = 0, because then you can't "divide them out" any more. (With thanks to dr. Seuss)

7. Apr 8, 2015

### RyanTAsher

Yes, the Bx is a B with subscript x, Bx, and the force value did have the $\hat k$ with it. Sorry for the confusion. I am pretty positive about the $\vec F = q \left( \vec B \times \vec v \right)$ as that is what it says in my textbook.

8. Apr 8, 2015

### BvU

The Lorentz force is definitely $$\vec F = q\, \left (\vec E + \vec v \times \vec B \right )$$no matter what your textbook says.