91Ó°ÊÓ

The Earth's magnetic field is a fascinating and critical component of our planet. It acts like a giant magnet with a North and South pole. Particularly, at any given location on Earth, the magnetic field has both a vertical and horizontal component. The field lines run from the magnetic North to the magnetic South. In many situations, such as navigating with a compass, we are primarily concerned with the horizontal component.

The horizontal component of the Earth's magnetic field is responsible for directing the needle of a compass. In our exercise, this component is due north, meaning that it points along the imaginary geographic meridian line towards the North Pole. It's important to note that this field is generally quite weak, typically measured in microteslas (µT).

• The horizontal magnetic field strength at the given location is provided as: \(2.5 \times 10^{-5} \text{ T}\)
• Horizontal component plays a crucial role in affecting charged particles moving on Earth's surface.

Proton motion under the influence of a magnetic field is an intriguing concept. Charged particles, like protons, experience a magnetic force when they move through a magnetic field. This force is always perpendicular to both the velocity of the particle and the magnetic field direction.

For a proton moving eastward, as in this exercise, the magnetic force can be calculated using the Lorentz force equation: \( F_{\text{magnetic}} = q \times v \times B \). Here, \( q \) is the charge of the proton (\(1.6 \times 10^{-19} \text{ C}\)), \( v \) is its velocity, and \( B \) is the magnetic field strength.

• The charge of a proton is positive, which is crucial for determining the direction of the magnetic force.
• When the proton moves eastward (to the right), with the field north (upward), the force acts vertically, affecting the force balance.
• This perpendicular nature of the force causes the proton's path to potentially curve in the absence of balanced forces.

Force balance is a fundamental concept in physics where two or more forces acting on an object are simultaneously in equilibrium, canceling each other out. In this scenario, the proton experiences both its weight, due to gravity, and the magnetic force from the Earth's magnetic field.

The weight of the proton is calculated using: \( F_{\text{gravity}} = m_{\text{proton}} \times g \), where \( m_{\text{proton}} = 1.67 \times 10^{-27} \text{ kg}\) and \( g = 9.81 \text{ m/s}^2\). For the forces to be balanced, \( F_{\text{gravity}} = F_{\text{magnetic}}\). By equating these forces: \( m_{\text{proton}} \times g = q \times v \times B \).

To find the speed \( v \), rearrange the equation:
\[ v = \frac{m_{\text{proton}} \times g}{q \times B} \]

• Insert known values to find: \( v = \frac{(1.67 \times 10^{-27}) \times 9.81}{(1.6 \times 10^{-19}) \times (2.5 \times 10^{-5})} \)
• Compute to get \( v \approx 4.025 \times 10^{4} \text{ m/s} \)
• This speed ensures that the gravitational and magnetic forces precisely balance.