7/√2 units
15 units
Assuming we know the horizontal component and we want to relate the total intensity of the horizontal component of Earth's magnetic field at the poles to its value at the equator. We assume an idealized dipole model of Earth's magnetic field.
Earth's magnetic field can be approximated as a magnetic dipole located at the Earth's center.
At the magnetic equator, the magnetic field is horizontal, and its magnitude is given by:
\( B_{equator} = \frac{\mu_0}{4\pi} \frac{M}{r^3} \)
Where:
At the magnetic poles, the horizontal component of the magnetic field is considered to be zero since the magnetic field is mostly vertical. However, if we're considering the total *intensity* (magnitude) of the magnetic field at a point slightly off the pole (where there is a small horizontal component), the horizontal component near the poles can be approximated as:
\( B_{pole} = \frac{\mu_0}{4\pi} \frac{M}{r^3} \)
*The reason that the formula near pole and the equator is so close is because that is the intensity of the horizontal component. We make a change to pole's formula: \( B_{pole} = \frac{\mu_0}{4\pi} \frac{M \cos \theta}{r^3} \)
*Therefore to get that relationship, we take the ratio \( B_{pole}/ B_{equator} = {\sqrt{2}}\) or \( B_{equator} = B_{pole}/{\sqrt{2}}\)
Given total horizontal intensity near the poles = 7 units \( B_{equator} = 7/{\sqrt{2}}\)
Given the total intensity of the *horizontal component* of the magnetic field near the poles to be 7 units, the total horizontal intensity at the equator is approximately \(\frac{7}{\sqrt{2}}\) units.
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: