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 tangential acceleration of the protons in meters/second^2: positive if they speed up and negative if they slow down.
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To calculate the tangential acceleration of the protons, we need to consider the concepts of circular motion and the relationship between acceleration and magnetic fields.
Here's how to approach the problem step by step:
1. Recall that the force experienced by a charged particle moving in a magnetic field is given by the equation:
F = q * v * B * sin(θ)
Where:
- F is the force experienced by the particle,
- 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.
2. In this case, we are interested in the tangential acceleration of the protons. Tangential acceleration is related to the force acting in the same direction as the velocity. So we need to find the component of the force in the tangential direction.
3. Since the path is circular, the velocity vector is always perpendicular to the radius vector. Therefore, the angle θ between the velocity vector and the magnetic field vector is 90 degrees.
4. Substituting the known values and simplifying, the force can be expressed as:
F = q * v * B
5. In circular motion, the force required to keep an object moving in a circular path is given by:
F = m * a
Where:
- m is the mass of the particle,
- a is the acceleration.
6. Equating the two expressions for force (from step 4 and step 5), we have:
q * v * B = m * a
7. Canceling out the charge (q) and rearranging the equation, we can solve for the tangential acceleration (a):
a = (q * v * B) / m
Now, let's crunch the numbers:
- The charge of the proton, q, is known to be approximately 1.6 * 10^-19 Coulombs.
- The mass of the proton, m, is roughly 1.67 * 10^-27 kilograms.
- The initial velocity of the protons can be assumed to be constant, as the question does not mention any changes in velocity.
- The rate of change of the magnetic field strength is given as 0.001 Tesla/second.
- The radius of the circular path is given as 1 meter.
Note: The velocity of the protons is not given directly, but it is useful to know that in circular motion, the speed of an object moving in a circle is given by v = ω * r, where ω is the angular velocity. The angular velocity is the rate at which the proton moves around the circle measured in radians/second. With this information, we can find v.
To obtain the tangential acceleration (a), substitute the known values into the equation above, and solve for a.