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3 years ago

10

The vectors of the magnetic field around a long, straight, current-carrying wire are:

2 answers:

Ivahew [28]3 years ago

6 0

Answer:

concentric

Explanation:

odyssey ware says correct

Ksenya-84 [330]3 years ago

3 0

Divide the long wire in small segments. The field due to one segment can be found from Biot- Savart law.

dB=(uo/4pi) Idlxr/r^3. In this expression if we take vector r perpendicular to current element , the cross product in the expression shows that the direction of B is perpendicular to vector r and from symmetry it can be said that field line is circular around the wire segment. This is true for all segment ,because in infinitely long wire all segments are equivalent . Thus, magnetic field lines are concentric around a long straight wire.

Hope it helps

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A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon

Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

$n=\frac{m}{M_m}$

where

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m = 1.4 mg = 0.0014 g is the mass of mercury

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Substituting, we find

$n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol$

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

$N=n N_A$

where

n is the number of moles

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Substituting,

$N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18}$ atoms

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$E_1 = \frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

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Substituting,

$E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J$

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4 0

3 years ago

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patriot [66]

Answer:

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Solution:

As per the question:

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Now,

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By using the formula:

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where

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Thus

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5 0

3 years ago