From the surface of a cylindrical wire of radius R=1 cm carrying a current I=1 mA an electron escapes with velocity v0=30 m/s perpendicular to the surface of the wire. What will be the electron's maximum distance in meters from the axis of the wire?
Details and assumptions
μ04π=10−7 H/m
e/me=1.76×1011 C/kg
Well, that electron seems like it really wanted to escape the constraints of that wire! Quite determined, I must say.
To find the maximum distance the electron will reach from the axis of the wire, we can use some good ol' physics. We know that the magnetic force on a moving charged particle is given by the equation F = qvB, where F is the force, q is the charge, v is the velocity, and B is the magnetic field.
Since the electron is moving in a direction perpendicular to the surface of the wire, the magnetic force will act as the centripetal force required to keep the electron on its circular path. We can write this as F = mv²/r, where m is the mass of the electron and r is the distance from the axis of the wire.
Now, setting the magnetic force equal to the centripetal force, we have qvB = mv²/r. Rearranging this equation, r = mv/(qB).
Plugging in the given values and constants, we have r = (9.1 x 10^(-31) kg)(30 m/s)/((1.76 x 10^(-19) C)(10^(-7) H/m)).
Calculating that out gives us r = approximately 3.85 x 10^(-3) meters.
So, the electron's maximum distance from the axis of the wire will be about 3.85 millimeters.
Just remember, electrons are quite the escape artists. Keep an eye on them, especially when there's a current involved!
To find the maximum distance of the electron from the axis of the wire, we can use the magnetic field created by the current in the wire.
The magnetic field at a distance r from the axis of a cylindrical wire carrying a current I is given by the formula:
B = μ₀I/2πr
Where:
B is the magnetic field
μ₀ is the permeability of free space (μ₀ = 4π × 10^(-7) H/m)
I is the current
r is the distance from the axis of the wire
Given:
R = 1 cm = 0.01 m (radius of the wire)
I = 1 mA = 0.001 A (current flowing through the wire)
v₀ = 30 m/s (velocity of the electron)
First, let's calculate the magnetic field at the surface of the wire (r = R):
B = μ₀I/2πR
Substituting the values:
B = (4π × 10^(-7) H/m)(0.001 A) / (2π × 0.01 m)
B = 2 × 10^(-5) T
Now, we know that the magnetic force experienced by the electron moving perpendicular to the magnetic field is given by the formula:
F = q * v * B
Where:
F is the force
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field
The charge of an electron is -e (e = 1.76 × 10^11 C/kg), but since we are only interested in the magnitude of the force, we can use the value e instead.
F = e * v * B
Substituting the values:
F = (1.76 × 10^(-11) C)(30 m/s)(2 × 10^(-5) T)
F = 1.056 × 10^(-9) N
The force acting on the electron provides the centripetal force to keep it in a circle around the wire. The centripetal force is given by the formula:
F = m * (v^2 / r)
Where:
m is the mass of the particle
v is the velocity of the particle
r is the radius of the circular path
The mass of an electron is me = 9.11 × 10^(-31) kg.
We can set these two formulas for force equal to each other:
m * (v^2 / r) = e * v * B
Simplifying the equation:
m * v / r = e * B
Solving for r:
r = m * v / (e * B)
Substituting the values:
r = (9.11 × 10^(-31) kg)(30 m/s) / ((1.76 × 10^(-11) C)(2 × 10^(-5) T))
r ≈ 7.68 × 10^(-3) m
Therefore, the electron's maximum distance from the axis of the wire is approximately 7.68 × 10^(-3) meters.
To answer this question, we need to use the concepts of magnetic fields and the Lorentz force experienced by a charged particle moving in a magnetic field.
First, let's determine the magnetic field at the surface of the wire. The magnetic field generated by a current-carrying wire is given by the formula:
B = (μ₀ * I) / (2π * R),
where B is the magnetic field, μ₀ is the permeability of free space (μ₀ = 4π * 10^(-7) H/m), I is the current, and R is the radius of the wire.
Given that the radius of the wire is 1 cm (0.01 m) and the current is 1 mA (0.001 A),
B = (4π * 10^(-7) H/m * 0.001 A) / (2π * 0.01 m)
= (4 * 10^(-7) * 0.001 A) / 0.02
= 0.2 * 10^(-7) T
= 2 * 10^(-8) T.
Next, let's consider the Lorentz force experienced by the electron when it escapes from the wire. The Lorentz force equation is given by:
F = q * (v x B),
where F is the force, q is the charge of the particle, v is the velocity vector, and B is the magnetic field vector.
The force experienced by the electron will be perpendicular to both its velocity and the magnetic field, causing it to move in a circular path.
The magnitude of the force experienced by the electron can be written as:
F = |q| * |v| * |B| * sin(θ),
where θ is the angle between the velocity vector and the magnetic field vector.
Since the electron escapes perpendicular to the surface of the wire, the angle between the velocity and the magnetic field is 90 degrees. Therefore, sin(θ) = 1.
Hence, F = |q| * |v| * |B|.
Now, let's solve for the distance from the wire's axis at which the electron will reach its maximum distance. At the point of maximum distance, the force experienced by the electron will be equal to the centripetal force.
The centripetal force is given by:
F_c = (m * v₀^2) / r,
where F_c is the centripetal force, m is the mass of the electron (m = e / (e / me)), v₀ is the velocity of the electron at the point it escapes, and r is the maximum distance from the wire's axis.
Setting F equal to F_c and solving for r:
m * v₀^2 / r = |q| * |v| * |B|,
r = (m * v₀^2) / (|q| * |v| * |B|).
Plugging in the given values:
m = (e / (e / me)) = (1.76 * 10^(11) C/kg) / (1.6 * 10^(-19) C) * (9.11 * 10^(-31) kg) ≈ 10^-3 kg,
v₀ = 30 m/s,
q = e = 1.6 * 10^(-19) C,
|v| = v₀ = 30 m/s,
|B| = 2 * 10^(-8) T,
r = ((10^(-3) kg) * (30 m/s)^2) / ((1.6 * 10^(-19) C) * (30 m/s) * (2 * 10^(-8) T))
= (10^(-3) kg * 900 m^2/s^2) / (1.6 * 10^(-19) C * 60 m^2/s^2)
= (9 * 10^(-4) kg * m^2/s^2) / (1.6 * 10^(-19) C * m^2/s^2)
= (9/16) * 10^(-23) m^2/C
≈ 5.625 * 10^(-23) m.
Therefore, the electron's maximum distance from the axis of the wire will be approximately 5.625 * 10^(-23) meters.