In earlier learning sequences we described how a static magnetic field cannot change the speed (and therefore kinetic energy) of a free charged particle. A changing magnetic field can, and this is one way particle beams are accelerated. Consider free protons following a circular path in a uniform magnetic field with a radius of 1 meter. At t = 0, the magnitude of the uniform magnetic field begins to increase at 0.001 Tesla/second. Enter the the acceleration of the protons in meters/second^2: positive if they speed up and negative if they slow down.
To find the acceleration of the protons, we need to use the equation for the centripetal acceleration of a charged particle moving in a magnetic field.
The equation is given by:
a = (q * v * B) / m
Where:
a is the acceleration
q is the charge of the particle (in this case, the charge of a proton)
v is the velocity of the particle
B is the magnetic field strength
m is the mass of the particle (in this case, the mass of a proton)
Given that the protons are following a circular path with a radius of 1 meter, we can determine the velocity of the protons using the formula for the centripetal force:
F = m * v^2 / r
Where:
F is the centripetal force
m is the mass of the proton
v is the velocity of the proton
r is the radius of the circular path
Since the force is provided by the magnetic field according to the equation F = q * v * B, we can equate the two equations:
q * v * B = m * v^2 / r
Simplifying the equation, we get:
v = (q * B * r) / m
Now, differentiate both sides with respect to time (t) to get the expression for the rate of change of velocity (dv/dt):
dv/dt = (dq/dt * B * r) / m
Substituting the given rate of change of magnetic field (0.001 Tesla/second), we have:
dv/dt = (0.001 * e * B * r) / m
Where e is the elementary charge of a proton.
Now, we know that acceleration is the rate of change of velocity (dv/dt). Therefore, the acceleration of the protons can be given by:
a = dv/dt = (0.001 * e * B * r) / m
Substituting the values:
- e = 1.602 x 10^-19 C (elementary charge)
- B = 0.001 Tesla (rate of change of magnetic field)
- r = 1 meter (radius of the circular path)
- m = 1.6726219 x 10^-27 kg (mass of a proton)
We can calculate the acceleration as follows:
a = (0.001 * 1.602 x 10^-19 C * 0.001 Tesla * 1 meter) / (1.6726219 x 10^-27 kg)
a = 0.0009636 m/s^2
Therefore, the acceleration of the protons is approximately 0.0009636 meters/second^2.
To determine the acceleration of the protons, we need to use the equation for the magnetic force acting on a charged particle moving in a magnetic field.
The equation for the magnetic force (F) on a charged particle is given by:
F = qvBsinθ
where:
- F is the magnetic force
- q is the charge of the particle
- v is the velocity of the particle
- B is the magnetic field strength
- θ is the angle between the velocity vector and the magnetic field vector
In this case, the protons are moving in a circular path, so the velocity (v) is tangential to the circle. This means that the angle θ between the velocity and the magnetic field is 90 degrees, and sinθ = 1.
The magnetic force (F) can be rearranged to solve for the acceleration (a):
F = ma
where:
- m is the mass of the particle
Since the protons have a constant mass, we can rewrite the equation as:
a = F/m
Now, let's calculate the acceleration:
Given:
- Radius of the circular path (r) = 1 meter
- Rate of change of the magnetic field (dB/dt) = 0.001 Tesla/second
- Charge of a proton (q) = 1.602 x 10^-19 Coulombs
- Mass of a proton (m) = 1.6726 x 10^-27 kilograms
The magnetic field strength (B) at any point on the circular path is given by:
B = μ0I/2πr
where:
- μ0 is the permeability of free space (4π x 10^-7 Tm/A)
- I is the current flowing through the circular path
Since the protons are not connected to a wire and are moving freely, there is no current flowing. Therefore, the magnetic field strength is zero.
Substituting the values into the acceleration equation:
a = (qvBsinθ) / m
= (1.602 x 10^-19 C) * (0 T) * (1)
= 0
Therefore, the acceleration of the protons in this scenario is 0 meters/second^2. They neither speed up nor slow down.