A coaxial cable consists of a thin insulated straight wire carrying a current of 2.00 A surrounded by a cylindrical conductor carrying a cur

Question

A coaxial cable consists of a thin insulated straight wire carrying a current of 2.00 A surrounded by a cylindrical conductor carrying a current of 3.50 A in the opposite direction. The cylindrical conductor has a radius of 0.420 cm. What is the magnitude of the magnetic field outside of the cylindrical conductor 2.00 cm from the central wire

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Thanh Thu 3 years 2021-08-26T08:18:44+00:00 1 Answers 25 views 0

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  1. bonexptip
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    2021-08-26T08:20:26+00:00
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    Answer:

    B = 15μT

    Explanation:

    In order to calculate the magnitude of the magnetic field generated by the coaxial cable you use the Ampere’s law, which is given by:

    B=\frac{\mu_oI}{2\pi r}       (1)

    μo: magnetic permeability of vacuum = 4π*10^-7 T/A

    I: current

    r: distance from the wire to the point in which B is calculated

    In this case you have two currents with opposite directions, which also generates magnetic opposite magnetic fields. Then, you have (but only for r > radius of the cylindrical conductor) the following equation:

    B_T=B_1-B_2=\frac{\mu_o I_1}{2\pi r}-\frac{\mu_o I_2}{2\pi r}\\\\B_T=\frac{\mu_o}{2\pi r}(I_1-I_2)  (2)

    I1: current of the central wire = 2.00A

    I2: current of the cylindrical conductor = 3.50A

    r: distance = 2.00 cm = 0.02 m

    You replace the values of all parameters in the equation (2), and you use the absolute value because you need the magnitude of B, not its direction.

    |B|=|\frac{4\pi*10^{-7}T/A}{2\pi (0.02m)}(2.00A-3.50A)|=1.5*10^{-5}T\\\\|B|=15*10^{-6}T=15\mu T

    The agnitude of the magnetic field outside the coaxial cable, at a distance of 2.00cm to the center of the cable is 15μT

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