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When we discuss electron acceleration in the context of magnetic fields, we are essentially focusing on how an electron's motion is influenced by these magnetic forces. In this scenario, a long, straight wire carrying a current creates a magnetic field around it. This field interacts with an electron that moves parallel to the wire. The solution to determining the electron's acceleration starts with calculating the magnetic field strength using the formula:
\( B = \frac{\mu_0 I}{2\pi r} \),
where \( \mu_0 \) is the permeability of free space. Once the magnetic field \( B \) is determined, the force \( F \) on the electron can be calculated using the formula:
\( F = qvB \sin \theta \).Knowing that \( \theta \) is \( 90^\circ \) because the electron is moving parallel to the current, \( \sin \theta = 1 \). With this force, the acceleration \( a \) of the electron is found using Newton's second law:
\( a = \frac{F}{m} \).Here, \( m \) is the mass of the electron. The result is a large acceleration value, indicating the significant impact of magnetic forces on the electron's motion.

The electric field's role is to counteract the magnetic force acting on an electron, allowing it to continue moving without any deviation from its path. To achieve this balance, the electric field must exert a force that is equal in magnitude and opposite in direction to the magnetic force already calculated. We establish this balance through the relation:
\( F_E = eE \),
where \( F_E \) is the electric force, and \( E \) is the electric field. By setting \( F_E = F_B \), the magnetic force, we solve for the electric field:
\( E = \frac{F_B}{e} \).The computed electric field must be radially outward from the wire to oppose the inward magnetic force, ensuring the electron travels parallel to the wire without experiencing a net force that would alter its trajectory.

Calculating forces acting on a moving charge is crucial to understanding the interaction of magnetic and electric fields. In this exercise, we focus on the magnetic force \( F \) calculated using the Lorentz force law, which is expressed as:
\( F = qvB \sin \theta \).Given the conditions of the problem, where an electron with charge \( q = -1.6 \times 10^{-19} \) C moves with velocity \( v \) parallel to the wire, the angle \( \theta = 90^\circ \) ensures that the magnetic force attains its maximum value. This maximum is because \( \sin 90^\circ = 1 \), simplifying our equation.Furthermore, to validate the need for an electric field counteracting this force, we compute:
\( F_B = F_E = eE \).Such computations underline how both electric and magnetic forces can be harmonized to ensure that the electron's trajectory remains unaffected by these fields.

In high-speed particle motion, such as electrons influenced by magnetic and electric fields, gravitational forces can often be neglected due to their negligible magnitude. For the electron in this exercise, the gravitational force \( F_g \) is calculated as:
\( F_g = mg \),
with \( m = 9.11 \times 10^{-31} \) kg and \( g = 9.8 \) m/s\(^2\). The resulting force \( F_g \approx 8.93 \times 10^{-30} \) N pales in comparison to the calculated magnetic force \( F_B \approx 5.2 \times 10^{-18} \) N.This immense difference underscores why gravitational effects are often discarded in scenarios involving strong electromagnetic forces compared to the relatively weak gravitational force on electrons. It reminds us of the relative scales of forces in physics and helps focus on significant effects for practical problem-solving.