An electron moving through a uniform magnetic field

In summary, the problem involves an electron moving through a uniform magnetic field with given velocity and magnetic force. The equation used is Fb = q(B x V), and after using the cross product resultant, the equation becomes (2.43 x 10^-19)k = (-3.30 x 10^-19*Bx) k. To solve for Bx, the k vectors can be cancelled out or the dot product can be taken.
  • #1
Destroxia
204
7

Homework Statement



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. Homework Equations


Fb = q(B x V)

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?
 
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  • #2
RyanTAsher said:

Homework Statement



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. Homework Equations


Fb = q(B x V)

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?
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
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
BvU said:
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.
I believe Ryan is referring to the two k vectors.
 
  • #5
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 [itex] \vec B [/itex] is uniform and not a function of x.) Is that correct?

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

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

In the initial problem statement, the force was given as (2.43 × 10-19) N with no direction specified. Later, you tacked on a [itex] \hat k [/itex]. I'm just wondering if the original problem statement had the [itex] \hat k [/itex] 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:
  • #6
@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
collinsmark said:
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 [itex] \vec B [/itex] is uniform and not a function of x.) Is that correct?

Are you sure about [itex] \vec F = q \left( \vec B \times \vec v \right) [/itex]? Is that correct or is something backwards?
I'm just wondering if the original problem statement had the [itex] \hat k [/itex] in it, or if the (2.43 × 10-19) N figure was meant to be magnitude only without specifying a direction.

Yes, the Bx is a B with subscript x, Bx, and the force value did have the [itex] \hat k [/itex] with it. Sorry for the confusion. I am pretty positive about the [itex] \vec F = q \left( \vec B \times \vec v \right) [/itex] as that is what it says in my textbook.
 
  • #8
The Lorentz force is definitely $$
\vec F = q\, \left (\vec E + \vec v \times \vec B \right )
$$no matter what your textbook says.
 

Related to An electron moving through a uniform magnetic field

What is a uniform magnetic field?

A uniform magnetic field is a region in space where the strength and direction of the magnetic field are constant. This means that a charged particle moving through the field will experience the same force at all points within the field.

What is an electron?

An electron is a subatomic particle that carries a negative charge. It is one of the fundamental particles that make up atoms, along with protons and neutrons. Electrons are found in the electron cloud surrounding the nucleus of an atom.

How does an electron move through a uniform magnetic field?

When an electron moves through a uniform magnetic field, it experiences a force due to the interaction between its negative charge and the magnetic field. This force causes the electron to move in a circular path perpendicular to the direction of the magnetic field.

What is the direction of the force on an electron moving through a uniform magnetic field?

The direction of the force on an electron moving through a uniform magnetic field is perpendicular to both the direction of the electron's velocity and the direction of the magnetic field. This is known as the right-hand rule.

How does the speed of an electron affect its motion in a uniform magnetic field?

The speed of an electron does not affect the radius of its circular motion in a uniform magnetic field. However, a higher speed will result in a larger frequency of revolution, meaning the electron will complete more revolutions per second.

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