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Setler [38]
3 years ago
8

In 1962 measurements of the magnetic field of a large tornado were made at the Geophysical Observatory in Tulsa, Oklahoma. If th

e magnitude of the tornado's field was B = 1.55 ✕ 10−8 T pointing north when the tornado was 8.55 km east of the observatory, what current was carried up or down the funnel of the tornado? Model the vortex as a long, straight wire carrying a current.
Physics
1 answer:
atroni [7]3 years ago
5 0

Answer:

Explanation:

we have:

B = 1.55 \times  10^{-8} T\\r = 8.55 km= 8.55 x 10^3m

We know:

B= \frac{\mu_0I}2\pi r}\\\\\therefore I =\frac{2\pi rB}{\mu_0}... (\mu_0 = 4\pi \times 10^{-7} T-m/A)\\\\= \frac{[2\pi ( 8.55 \times 10^3m)(1.55 \times 10^{-8} T)]}{4\pi \times 10^{-7} T-m/A}\\\\ =662.625 A

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Answer:

Total 3 holes are available for conduction of current at 300K.

Explanation:

In order to develop a semiconductor, two type of impurities can be added as given below:

  1. N-type Impurities: Pentavalent impurities e.g. Phosphorous, Arsenic are added to have an additional electron in the structure. Thus a pentavalent impurity creates 1 additional electron.
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Now for estimation of extra electrons in the impured structure is as

N_{electrons-free}=n_{pentavalent \, atoms}\\N_{electrons-free}=4\\

Now for estimation of "holes"  in the impured structure is as

N_{holes}=n_{trivalent \, atoms}\\N_{holes}=7\\

Now when the free electrons and "holes" are available in the structure ,the "holes" will be filled by the free electrons therefore

N_{holes-net}=N_{holes}-N_{electrons-free}\\N_{holes-net}=7-4\\N_{holes-net}=3

So total 3 "holes" are available for conduction of current at 300K.

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