Vector evaluation & Lorentz force law

[Roodles01]

Homework Statement

An electron in a magnetic field B=2.0T(e_x-e_z) has velocity v=(2.5x10⁷ ms-1 (e_x-e_y)
a) calculate magnetic force on electron at that instant
b) what is the magnitude of this force

Homework Equations

I am using the Lorentz force law.
F = q(vXB)
& evaluating the directions by vector

My argument is that the e_y term is negative, but another person evaluates this as positive. Please confirm whether I'm right or wrong.

This affects the outcome of magnitude of force, too.

The Attempt at a Solution

F = -e 2(2.5x10⁷) (e_x-e_z)* (e_x-e_y)

I ex ey ez I
I 1 . -1 . 0 I
I 1 . 0 . -1 I

= ex((-1x-1)-(0x0)) + ey((1X-1)-(0x1)) + ez ((1x0)-(-1x1))
= e_x - e_y + e_z)

so
F = -8x10-12 (e_x - e_y + e_z)

 

[rude man]

Better believe your friend about the sign of the j term! (Re-check your determinant).

 

[Roodles01]

Ah! Having looked at it, I found that I didn't change the sign for the second term.
So simple it's annoying.
Thank you.

 

[rude man]

Ah! Having looked at it, I found that I didn't change the sign for the second term.
So simple it's annoying.

Welcome to the club!

 

[Nickie]

= (-8x10-12, 8x10-12, 0)

a) To calculate the magnetic force on the electron at this instant, we can use the Lorentz force law, which states that the force on a charged particle moving through a magnetic field is given by the equation F = q(v x B), where q is the charge of the particle, v is its velocity, and B is the magnetic field. In this case, the electron has a charge of -e and is moving with a velocity of 2.5x107 m/s in the direction of (ex-ey). The magnetic field is given as B=2.0T(ex-ez). Plugging these values into the equation, we get:

F = -e(2.5x107)(ex-ey) x (2.0T(ex-ez))

Using the cross product, we can find the direction of the force:

(ex-ey) x (ex-ez) = ex((-1x-1)-(0x0)) + ey((1x-1)-(0x1)) + ez((1x0)-(-1x1)) = ex - ey + ez

So, the force on the electron is given by:

F = -e(2.5x107)(ex - ey + ez) = (-8x10-12, 8x10-12, 0) N

b) The magnitude of this force can be found using the Pythagorean theorem:

|F| = √((-8x10-12)2 + (8x10-12)2 + 02) = 1.13x10-11 N

Therefore, the magnitude of the magnetic force on the electron is 1.13x10-11 N.

In regards to the direction of the force, both the x and y components have a negative value, while the z component is zero. This means that the force is in the direction of the negative x and y axes, towards the negative yz plane. Whether the ey term is positive or negative does not affect the magnitude of the force, as it will just change the direction of the force vector. In this case, it seems that you have correctly evaluated the direction of the force. It is important to keep in mind that when using vector equations, the direction of the vectors is crucial and a small mistake in direction can greatly affect the outcome.