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PilotLPTM [1.2K]

2 years ago

9

On isolated ground receptacles, the metal yoke ____ integrally bonded to the equipment grounding terminal of the receptacle.

1 answer:

Ede4ka [16]2 years ago

4 0

On isolated ground receptacles, the metal yoke is not allowed to be integrally bonded to the equipment grounding terminal of the receptacle.

Any device with two distinct switches or receptacles is a duplex device. It can be shaped to fit a Decora opening or a typical duplex plate opening. It should be noted that they can be combination devices with a switch/outlet, switch/pilot light, etc.

Because of grounding connection removal and receptacle, it is utterly undesirable to connect the two bare equipment grounding conductors together directly.

The equipment grounding conductor associated with those circuits must be connected to the box when circuit conductors are terminated on equipment inside a metal box to prevent unneeded current discharge.

Learn more about grounding conductors here brainly.com/question/14886979

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jeka57 [31]

Answer:

$r = 3.36 \times 10^{-7} m$

Explanation:

As per Ampere's law of magnetic field we know that

line integral of magnetic field along closed ampere's loop is equal to the product of current enclosed and magnetic permeability of medium

So it is given as

$\int B. dl = \mu_0 i_{en}$

here we can say that enclosed current is given as

$i_{en} = \frac{i}{\pi R^2} (\pi r^2)$

now from ampere'e loop law for any point inside the wire we will have

$B.(2\pi r) = \mu_o (\frac{ir^2}{R^2}$

$B = \frac{\mu_0 i r}{2\pi R^2}$

now we know that magnetic field inside the wire is 84% of the field at its surface

so we will have

$0.84 \frac{\mu_o i}{2\pi R} = \frac{\mu_o i r}{2\pi R^2}$

so we have

$r = 0.84 R$

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$\frac{\mu_o i}{2\pi R} = 1$

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$r = 3.36 \times 10^{-7} m$

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