You create a plot of voltage (in V) vs. time (in s) for an RC circuit as the capacitor is charging, where V=V_{0} \cdot \left(1-
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Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:
1)
When both the electric field and the magnetic field are acting on the electron normal to the beam and normal to each other, the electric force and the magnetic force on the electron have opposite directions: in order to produce no deflection on the electron beam, the two forces must be equal in magnitude
where
q is the electron charge
E is the magnitude of the electric field
v is the electron speed
B is the magnitude of the magnetic field
Solving the formula for v, we find
2) 4.1 mm
When the electric field is removed, only the magnetic force acts on the electron, providing the centripetal force that keeps the electron in a circular path:
where m is the mass of the electron and r is the radius of the trajectory. Solving the formula for r, we find
3)
The speed of the electron in the circular trajectory is equal to the ratio between the circumference of the orbit, , and the period, T:
Solving the equation for T and using the results found in 1) and 2), we find the period of the orbit: