Question

The magnetic dipole moment of the Earth is approximately \(8.0 \cdot 10^{22} \mathrm{~A} \mathrm{~m}^{2}\). The source of the Earth's magnetic field is not known; one possibility might be the circulation culating ions move a circular loop of radius \(2500 \mathrm{~km}\). What “current" must they produce to yield the observed field?

Short answer

Answer: The ions must produce a current of approximately \(4.08\times10^7~\text{A}\) to yield the observed magnetic field of Earth.

Step-by-step solution

Write down the given values.

The magnetic dipole moment of the Earth is given as \(8.0\times10^{22}~\text{A}~\mathrm{m}^{2}\). The radius of the circular loop is given as \(2500~\text{km}\).

Convert the radius to meters.

To work with SI units, we need to convert the radius from kilometers to meters: $$r = 2500~\text{km} \times 1000~\mathrm{m}/\text{km} = 2.5\times10^{6}~\text{m}$$

Express the magnetic dipole moment formula for a circular loop.

The formula for the magnetic dipole moment (\(\mu\)) of a circular loop with a current \(I\) and a loop radius \(r\) is given by: $$\mu = I \times \pi r^2$$

Solve for current I.

We can rearrange the formula for the magnetic dipole moment to solve for the required current: $$I = \frac{\mu}{\pi r^2}$$

Substitute the given values and calculate the current.

Now, substitute the given values for \(\mu\) and \(r\) and calculate the current \(I\): $$I = \frac{8.0\times10^{22}~\text{A}~\mathrm{m}^{2}}{\pi (2.5\times10^6~\mathrm{m})^2} \approx 4.08\times10^7~\text{A}$$ So, the ions must produce a current of approximately \(4.08\times10^7~\text{A}\) to yield the observed magnetic field of Earth.

Key concepts

Earth's Magnetic Field

The Earth's magnetic field is a fascinating aspect of our planet. It's akin to an enormous bar magnet embedded within the Earth. This magnetic field is responsible for many functions, like guiding compasses, protecting us from solar wind, and contributing to the auroras. However, the source of this magnetic field isn't straightforward. Scientists hypothesize that it may originate from the Earth's outer core, which consists of liquid iron and nickel. These materials create a dynamic, constantly flowing current, contributing to the magnetic field through processes like the geodynamo effect. Understanding this dynamic helps us appreciate the magnetism we experience daily.

Current Calculation

Calculating current in electromagnetic contexts involves understanding the relationship between current, magnetic field strength, and other physical parameters. In exercises like the one provided, the hypothetical need is to find what kind of current flow could produce Earth's magnetic dipole moment. This involves using the magnetic dipole moment formula, \(\mu = I \times \pi r^2\). Rearranging the formula to solve for current \(I\), we find \(I = \frac{\mu}{\pi r^2}\). Here, \(I\) represents the equivalent current, \(\mu\) is the magnetic dipole moment, and \(r\) is the radius of the loop, meaning the sphere-like path on which ions circulate. This way, we mathematically express how much 'current' is needed to yield equivalent magnetic properties.

Circular Loop Radius

A circular loop radius is crucial in understanding magnetic phenomena like the Earth's magnetic field. The radius of such a loop represents half of the diameter of the path our circulating ions travel as imagined in the context of this exercise. When considering the Earth's case, a radius of \(2500~\text{km}\) signifies this path's extent, in which hypothetical charged particles could circulate to mimic Earth's magnetic field properties. It's essential to convert this radius into meters when performing calculations since standard SI units make equations consistent and reliable. Thus, knowing the radius in meters makes further analyses, like calculating current, straightforward.

SI Units Conversion

Conversion into SI units is a fundamental step in solving most physics problems. It ensures standardization across calculations and allows for clear, consistent communication of scientific ideas. In our scenario, converting the radius from kilometers to meters is a necessary step because the standard SI unit for length is meters. For example, converting \(2500~\text{km}\) into meters involves multiplying by \(1000\), resulting in \(2.5 \times 10^6~\text{m}\). This conversion process isn't just a convenience; it ensures that all components of an equation are compatible and align with the globally accepted measurement standards. This practice is a cornerstone of accurate scientific computation and communication.