What is the Magnetic Force on an Electron Moving Near a Current-Carrying Wire?

[cseil]

Homework Statement

There's a straight wire through which passes a current of 48.8A.
There's an electron moving with velocity 1.08x10^7 m/s at distance 5.20cm from the wire.

I have to calculate the magnetic force on the electron when:
a) the velocity is directed perpendicularly to the wire
b) the velocity is directed parallel to the wire
c) the velocity is perpendicular to the other two directions (a,b)

Homework Equations

Biot-Savart's law
$B = \frac{μ_0i}{2πr}$

$F = q\vec{v}× \vec{B}$

The Attempt at a Solution

B generated by the wire is 1.88x10^-4 T

The direction of B is perpendicular to i and the vector $\vec{d}$ (the distance between the wire and the electron). I considered the i going from left to right, so the vector B points inside the plane.

(a) Considering that, the force should be perpendicular to $\vec{v}$ and $\vec{B}$ so, with v directed to the wire, parallel to the wire but directed opposide to the current. The value is ok, I can calculate it, but I don't know if verse is right. My book says that the force is parallel and has the same verse of the current. I don't know why.

(b) If B points inside the plane, v is parallel to the wire, F points to the wire. It has the same value of (a).

(c) F is 0 because v is parallel to B.

 

[ehild]

You have done it well, but you are unsure about the signs.
To make sure about the direction of the force, set up the coordinate axes at the place of the electron, and determine the force vector from the vector product. $\vec B = B \hat y$ and $\vec F = q \vec v \times \vec B$. If the electron moves perpendicularly to the wire and radially outward, $\vec v = v \hat x$.
$\vec F = q \vec v \times \vec B= q v B \hat x\times \hat y= qvB \hat z$ Note that the charge of the electron is -e. What is the direction of the force?
What is the direction if the electron moves radially inward ?
What is the direction of the force if the electron moves in the same direction as the current?

 

[cseil]

Thank you for the answer, I've understood!

The electon have a perpendicular direction and it is going towards the wire.
So $\vec{v} = -v\hat{x}$
$\vec{B} = B\hat{y}$

$F = -qv\hat{x} \times B\hat{y}$

So considering that q is -e, I have

$F = evB \hat{z}$

The direction of the force is z, it is positive so in the case (a) the force is (as the book says) parallel to the wire and it has the same verse of the current.

So, case b.

I have $\vec{v} = v\hat{z}$

$F = -ev\hat{z} \times B\hat{y} = -evB\hat{x}$

The force is radial inward (direction x and opposite verse of the axis).

Is that right?

 

[ehild]

So, case b.

I have $\vec{v} = v\hat{z}$

$F = -ev\hat{z} \times B\hat{y} = -evB\hat{x}$

The force is radial inward (direction x and opposite verse of the axis).

Is that right?

No. $\hat z \times \hat y = - \hat x.$ (remember: $\hat x \times \hat y=\hat z$, $\hat y \times \hat z=\hat x$, $\hat z \times \hat x=\hat y$, and changing the order, the product changes sign.

 

[cseil]

No. $\hat z \times \hat y = - \hat x.$ (remember: $\hat x \times \hat y=\hat z$, $\hat y \times \hat z=\hat x$, $\hat z \times \hat x=\hat y$, and changing the order, the product changes sign.

Oh, right.
So same direction, opposite verse.

Thank you!
It is clear now! :)