A solenoid 20 cm long is constructed from 500 loops of wire and carries a current of 5 A as seen the diagram below. A tiny hole is drilled through the diameter of the solenoid in such a way that the current through the solenoid is not disrupted. If an electron flies through the hole with a speed of 5x10^6 m/s. how much force will the electron feel when it enters the solenoid? what will be the initial direction of that force?
To determine the force experienced by the electron when it enters the solenoid, we need to use the formula for the force on a charged particle moving through a magnetic field:
F = q * (v * B * sinθ)
where:
F is the force on the electron,
q is the charge of the electron (-1.6 x 10^-19 C),
v is the velocity of the electron (5 x 10^6 m/s),
B is the magnetic field strength inside the solenoid,
θ is the angle between the velocity vector and the magnetic field vector.
To find the magnetic field strength inside the solenoid, we can use the formula for the magnetic field caused by a solenoid:
B = μ₀ * n * I
where:
B is the magnetic field strength,
μ₀ is the permeability of free space (4π x 10^-7 T*m/A),
n is the number of turns per unit length (number of loops divided by length),
I is the current flowing through the solenoid (5 A).
Let's calculate step by step:
1. Calculate the magnetic field strength (B) inside the solenoid:
n = 500 / 0.2 m⁻¹ = 2500 m⁻¹ (since the solenoid is 20 cm long and has 500 loops)
B = (4π x 10^-7 T*m/A) * (2500 m⁻¹) * (5 A)
2. Calculate the force (F) on the electron:
F = (-1.6 x 10^-19 C) * (5 x 10^6 m/s) * (B) * (sinθ)
Now, since the electron is entering the solenoid, its velocity vector will be perpendicular to the magnetic field vector (θ = 90°). So, the angle between the velocity vector and the magnetic field vector is 90°, and sinθ = 1.
Therefore, the force (F) on the electron can be calculated as:
F = (-1.6 x 10^-19 C) * (5 x 10^6 m/s) * (B) * 1
Now, plug in the value of B into the equation to calculate the force.
To calculate the force experienced by the electron when it enters the solenoid, we need to use the magnetic field inside the solenoid and the velocity of the electron.
1. First, let's find the magnetic field inside the solenoid using the formula:
B = μ₀ * (n * I)
where B is the magnetic field, μ₀ is the permeability constant (4π × 10^-7 T·m/A), n is the number of loops per unit length, and I is the current through the solenoid.
Given that the solenoid has 500 loops and is 20 cm long (0.2 m), we can calculate the number of loops per unit length (n) as follows:
n = 500 / 0.2 = 2500 loops/m
Substituting the values into the formula, we get:
B = (4π × 10^-7) * (2500) * (5)
B ≈ 7.854 × 10^-3 T
So, the magnetic field inside the solenoid is approximately 7.854 × 10^-3 T.
2. To determine the force on the electron, we can use the formula for magnetic force on a moving charge:
F = q * v * B * sin(θ)
where F is the force, q is the charge of the electron (1.602 × 10^-19 C), v is the velocity of the electron (5 × 10^6 m/s), B is the magnetic field, and θ is the angle between the velocity vector and the magnetic field vector.
The initial direction of the force can be determined by the right-hand rule. When the velocity of the electron is in the same direction as the magnetic field, the force will be perpendicular to both and will be directed out of the page.
Substituting the values into the formula, we get:
F = (1.602 × 10^-19) * (5 × 10^6) * (7.854 × 10^-3) * sin(θ)
As the angle θ is not specified, we assume it to be 90 degrees since the force is perpendicular to the velocity vector.
F ≈ 0.0788 N
The force experienced by the electron when it enters the solenoid is approximately 0.0788 N, directed out of the page.