Cross Product application problem

In summary, the magnitude of the magnetic force on an electron is 1.28x10-13N and the direction is in the y-axis.
  • #1
Asheram0
5
0
Calculating magnitude and direction of an magnetic force on an electron

Homework Statement



Okay so the question says that the magnetic force (Vector)FM on a particle which is in a magnetic field is found by (Vector)FM = Vector I x Vector B.

Vector I = charge multiplied by the velocity of a charged particle
Vector B= the strength of the magnetic field (Measured in Tesla(T) ).

An electron accelerated from REST to the RIGHT, in a horizontal directed electric field.
The electron then leaves the electric field at a speed of 4.0 x 10^6, entering the magnetic field of magnitude 0.20 Tesla that is directed into the screen.

From the given information calculate the magnitude and direction of the magnetic force on the electron given that the charge on an electron is q= 1.6 x 10^-19.


Homework Equations



All right, since we have to solve this in the cross product method I'm quite lost.
I take it we want to end up with something like this: (ax, ay, az) and b = (bx, by, bz)
giving us a x b = = (aybz − azby , azbx − axbz , axby − aybx)


The Attempt at a Solution



So far i analyzed that

Vector I = charge of an electron (q = 1.6 x 10^-19) x velocity of a charged particle (4.06 x 10^6) (Since it says electron accelerates from rest to the right horizontally, i assume vector I is the x component?)

and for

Vector B = strength of the magnetic field in Tesla (0.20 T). (And since it says its directed into the screen, i assume its the z component? Right?)

So now we have

Vector I = 1.6x10^-19(4.0 x 10^6) = 6.4 x 10^-13
Vector B = 0.20

|I||B|sin90
|6.4x10^-13| x |0.20| sin 90 = 1.28 x 10^-13

So the magnitude is 1.28 x 10^-13 and the direction is in the y axis?
I defenitley think i hugely messed up somewhere, please help
 
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  • #2
Your calculations look fine to me. But you should take care when dealing with vectors and vector components to make sure that you specify directions completely. "in the y-axis" doesn't indicate if the direction is towards +y or -y. You should be able to write each vector in component form: ## x \hat{i} + y \hat{j} + z \hat{k}## .
 
  • #3
gneill said:
Your calculations look fine to me. But you should take care when dealing with vectors and vector components to make sure that you specify directions completely. "in the y-axis" doesn't indicate if the direction is towards +y or -y. You should be able to write each vector in component form: ## x \hat{i} + y \hat{j} + z \hat{k}## .

I very much appreciate your reply, but since my answer is a positive value would it be correct in saying its towards the +y?

So i would do
|Vector Ix| = |vector I|cos 90°


6.4x10^-13 cos 90
=0

0.20 cos 90

=0

Sorry I am confused?
 
  • #4
Asheram0 said:
I very much appreciate your reply, but since my answer is a positive value would it be correct in saying its towards the +y?
The sign of the result should depend upon your initial choice of axes, how the given values "lie" in that frame of reference, and any operations performed on those values. It's helpful to specify the coordinate system at the beginning and state the given values in terms of those coordinates. So for example, you might state that the +x-axis is chosen to coincide with the electron's direction of motion, and that the +z-axis is into the screen. The magnetic field is parallel to the z-axis and is directed towards +z. Or more concisely you could write:

Let: ##\vec{v} = 4.0x10^6 \hat{i}~m/s~~~~~and~~~~~ \vec{B} = 0.20 \hat{k}~T ##
So i would do
|Vector Ix| = |vector I|cos 90°


6.4x10^-13 cos 90
=0

0.20 cos 90

=0

Sorry I am confused?
If the result happened to have some odd direction that didn't coincide with any of the axes then you might want to write out the vector in ijk form. But if it coincides with an axis then it would be sufficient to specify which axis and which direction along that axis. For example, 1.28x10-13 N in the direction of the +y-axis.
 
  • #5
Thank you very very much, it helped a lot! The right hand rule could also be applied in this question!
Kudos man!
 

Related to Cross Product application problem

1. What is the purpose of using cross product in application problems?

The cross product is used to find the direction of a vector perpendicular to two given vectors. In application problems, it is commonly used to calculate torque, magnetic force, and angular momentum.

2. How do you solve a cross product application problem?

To solve a cross product application problem, you first need to determine the given vectors and their direction. Then, you can use the cross product formula (a x b = |a||b|sinθ) to calculate the magnitude and direction of the resulting vector.

3. What are some real-life examples of cross product application problems?

Cross product application problems are commonly used in physics and engineering. Some examples include calculating the torque on a wrench, the magnetic force on a current-carrying wire, and the angular momentum of a rotating object.

4. Can cross product be used in two-dimensional problems?

Yes, cross product can be used in both two-dimensional and three-dimensional problems. However, in two-dimensional problems, the resulting vector will always be perpendicular to the given vectors and will lie along the z-axis.

5. How is cross product different from dot product?

The dot product is used to calculate the scalar product of two vectors, while the cross product is used to calculate the vector product. The dot product results in a scalar quantity, while the cross product results in a vector quantity. Additionally, the dot product is commutative, while the cross product is anti-commutative.

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