的意义是什么积分的负面磁场(nT)随着时间的?- 江南体育网页版- - - - -地球科学堆江南电子竞技平台栈交换 最近30从www.hoelymoley.com 2023 - 07 - 09 - t01:35:48z //www.hoelymoley.com/feeds/question/9341 https://creativecommons.org/licenses/by-sa/4.0/rdf //www.hoelymoley.com/q/9341 4 的意义是什么积分的负面磁场(nT)随着时间的? KcFnMi //www.hoelymoley.com/users/7194 2016 - 12 - 25 - t17:57:35z 2017 - 06 - 12 - t15:18:13z < p > < a href = " https://i.stack.imgur.com/G0drq.png " rel = " nofollow noreferrer " > < img src = " https://i.stack.imgur.com/G0drq.png " alt = "在这里输入图像描述" > < / > < / p > < p >我有一个磁场随时间(轴)的情节(轴),一个测量每小时(Dst -干扰Storm-Time)。< / p > < p >,不时有一个负峰,这表明磁性风暴。它越下降更大的风暴。< / p > < p >积分的物理意义是什么(nT x h)吗? < / p > //www.hoelymoley.com/questions/9341/-/10568 # 10568 2 user2821回答的的意义是什么积分的负面磁场(nT)随着时间的? user2821 //www.hoelymoley.com/users/2536 2017 - 06 - 12 - t15:15:40z 2017 - 06 - 12 - t15:15:40z < p > < a href = " http://wdc.kugi.kyoto-u.ac.jp/dstdir/dst2/onDstindex.html " rel = " nofollow noreferrer " >扰动风暴时间指数< / >是削弱水平分量的测量地球磁场在磁干扰。抑郁往往两侧山峰,启动和结束这场风暴。< a href = " https://en.wikipedia.org/wiki/Disturbance_storm_time_index " rel = " nofollow noreferrer " > < / > Dst是发表的< a href = " http://wdc.kugi.kyoto-u.ac.jp/dstdir/ " rel = " nofollow noreferrer " > < / >京都大学和更多信息的方法和引用的web页面。< / p > < blockquote > < p > Dst指数代表了轴对称扰动磁场在地球表面上的赤道偶极子。Dst的主要干扰是负面的,即减少在地磁场。这些领域减少生产主要由赤道洋流系统在磁层,通常称为环电流。中性表电流在磁性层的尾巴使附近的一个小领域的贡献减少地球。积极的Dst的变化主要是由于磁层的压缩从太阳风压力增加。< / p > < /引用> < p > < em > Sugiura龟井静香,IAGA公报没有40 (1991)< / em > < / p > < p >措施提出偏离平均磁场强度,以SI单位< a href = " https://en.wikipedia.org/wiki/Tesla_(单位)”rel = " nofollow noreferrer " >特斯拉T < / >美元在时间T美元。在这种情况下,磁场很弱,因此nano前缀n通常使用美元。 Magnetic field strength can be derived from weight (in $kg$), electric charge (in Coulomb $C$) and time in second ($s$) as:

$T =\frac{kg}{C \times s}$

if you integrate with time (in seconds, $s$) you'll get for each unit:

$T \times s =\frac{kg}{C}$.

The relation of electric charge and weight is used as a measure of Radiation exposure, often given in röntgen ($R$) and related as:

$1 \frac{C}{kg} = 3876 R$

Technically, your integral will, therefore, be a measure of increased radiation exposure, kinetic energy released per unit mass, for the time you integrate over. You could convert it to röntgen ($2.58×10^{−4} C/kg = R$) but note that the calculated product describes the deviation, not the absolute ionization.

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