A long straight and thin wire has a charge Q per unit length. It is also carrying a current I. A point charge is moving with a velocity v parallel to the conductor, at a distance d from it. Find the velocity v, supposing a correct direction for the current and the current sign in the conductor. I am a bit lost in this one...
My approach so far:
The electric field E at a distance d, generated by the conductor is:
E = lambda/(2*pi*d*e0) (where lambda is Q [c/m]
I assumed that the charge is positive, so the direction of the field pointing outwards the conductor.
The magnetic field generated by the current at a distance d from the conductor is:
B = u0*I/(2*pi*d)
If:
(x axis and y axis)
o -> Point charge Q
-------------- ->Conductor
-> I
So, the magnetic field is pointing out of the sheet and the electric force on the point charge Q is in the +y direction.
Using Newtons law, and assuming that the charge is moving parallel to the conductor:
Q*lambda/(2*pi*e0*d) - Q*v*u0*I/(2*pi*d) = 0
therefore, the charge's velocity is:
v = lambda/(e0*u0*I)
Is this correct or am I mistaken? Please Help!!!
Your approach is correct, but there are a couple of errors in your calculations. Let's go through it step by step to find the correct expression for the velocity, v.
First, you correctly determine that the electric field, E, at a distance d from the wire is given by:
E = λ / (2πε₀d),
where λ is the charge per unit length and ε₀ is the permittivity of free space.
Next, you correctly determine that the magnetic field, B, produced by the current in the wire is given by:
B = μ₀I / (2πd),
where I is the current in the wire and μ₀ is the permeability of free space.
Now, when you consider the force acting on the charge Q due to both the electric and magnetic fields, you need to take into account both the magnitude and the direction of the forces.
The electric force, Felectric, acting on the charge Q is given by:
Felectric = QE,
where Q is the charge of the moving particle and E is the electric field. In this case, Q is positive, so the direction of the electric force is in the same direction as the electric field, i.e., in the +y direction.
The magnetic force, Fmagnetic, acting on the charge Q is given by:
Fmagnetic = QvB.
In this case, the charge Q is moving parallel to the conductor, so the direction of the velocity v is perpendicular to both the magnetic field B and the direction of the current. Applying the right-hand rule, you can determine that the magnetic force is in the +z direction.
Now, setting the sum of the electric and magnetic forces equal to zero, we have:
Felectric + Fmagnetic = 0.
Substituting the expressions for the forces and rearranging, we get:
QE + QvB = 0.
Now, substituting the expressions for the electric and magnetic fields, we obtain:
Q(λ / (2πε₀d)) + Qv(μ₀I / (2πd)) = 0.
Finally, rearranging and solving for v, we find:
v = - (λ / (ε₀μ₀I)).
In conclusion, the correct expression for the velocity v of the moving charge Q, as a function of the charge per unit length λ, the current I in the wire, and the constants ε₀ and μ₀, is:
v = - (λ / (ε₀μ₀I)).