A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3 mm and outer

Question

A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3 mm and outer radius 3.5 mm. A current of 15 A flows down the inner wire and an equal return current flows in the outer conductor. If we assume that the currents are uniform over the cross section of the conductors, then calculate the magnitude of the magnitude at a radius of 2.5 mm.

a. 2.4 mT
b. 2.1 uT
c. 3.2 uT
d. 1.2 mT
e. 0.

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Mít Mít 6 months 2021-08-15T00:51:05+00:00 1 Answers 35 views 0

Answers ( )

    0
    2021-08-15T00:52:34+00:00

    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

    B=\frac{\mu_0 I}{2\pi r}

    where

    \mu_0 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:

    B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT

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