The atomic number of an element is also the number of
1 answer:
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Answer:
The self-induced emf in this inductor is 4.68 mV.
Explanation:
The emf in the inductor is given by:
Where:
dI/dt: is the decreasing current's rate change = -18.0 mA/s (the minus sign is because the current is decreasing)
L: is the inductance = 0.260 H
So, the emf is:
Therefore, the self-induced emf in this inductor is 4.68 mV.
I hope it helps you!
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: