Electrons, wire, amps and magnetic fields?

[heston1980]

An electron is moving by a wire that is carrying 10 amps of current
e-----------------------------> (electrons)
0===============I-->======== (wire)

if the electron is moving by the wire at 2.5x10^6 m/s and is .33 m from the wire how do I find these two answers?

A) The magnetic field strength 0.33 m from the wire
B) The force exerted on the electron by the magnetic field? including the direction?

 

[da_nang]

An electron is moving by a wire that is carrying 10 amps of current
e-----------------------------> (electrons)
0===============I-->======== (wire)

if the electron is moving by the wire at 2.5x10^6 m/s and is .33 m from the wire how do I find these two answers?

A) The magnetic field strength 0.33 m from the wire
B) The force exerted on the electron by the magnetic field? including the direction?

A) What formula can you use to determine the strength of a magnetic field created by a current in a wire?

Spoiler

Biot-Savart's law.

B) What's the formula to determine the force exerted on a charged particle moving in a magnetic field?

Spoiler

Lorentz force.

 

[heston1980]

A) What formula can you use to determine the strength of a magnetic field created by a current in a wire?

Spoiler

Biot-Savart's law.

B) What's the formula to determine the force exerted on a charged particle moving in a magnetic field?

Spoiler

Lorentz force.

I'm totally lost and don't understand either formulas

 

[da_nang]

A current I that goes through a wire generates a magnetic field at distance r from the wire with a magnetic flux density B. The magnetic flux density is calculated as B = μ₀I/(2πr), where μ₀ is the magnetic constant. (π is pi) Use this for A).

In B), a charged particle with charge Q traveling at velocity v in a magnetic field with magnetic flux density B is exerted by the Lorentz force which is F = Q[E + vxB]. However, Lorentz force is sometimes used for the magnetic component. The magnetic force is expressed as F_m = QvxB (Italics indicate vectors, the x denotes the cross product).

In scalar form, this becomes F_m = QvBsin(θ), where θ is the angle between the velocity and magnetic field density vectors. Given that the particle moves parallell with the wire in B), the vectors are perpendicular and thus θ = 90º and sin(θ) = sin(90º) = 1. Thus F_m = QvB. Use this to calculate the force exerted on the electron.

The direction of the force and the magnetic field density vectors are given by the right-hand rule.

 

[heston1980]

F=1.602*10^-19*2.5*10^6*.33 ??

 

[da_nang]

F=1.602*10^-19*2.5*10^6*.33 ??

.33 is distance between the wire and the electron. What you want to have in its place is B = μ₀I/(2πr).

Basically, you have these data:

Q = -1.6021773*10^-19 C (Although in this assignemnt, you can skip the negative sign and find out the direction later.)
v = 2.5*10⁶ m/s
I = 10 A
r = 0.33 m

Now for A), plug the corresponding data into the formula B = μ₀I/(2πr). For B), to get a more precise value, substitute B in the formula for the magnetic force:

F = QvB = Qvμ₀I/(2πr). Plug in the data to get your answer.