91Ó°ÊÓ

In this exercise, we are dealing with a uniform magnetic field. That means the magnetic field has the same strength and direction at every point within a certain region. This uniformity simplifies calculations because we don't have to worry about variations in the field.

Imagine the magnetic field as a set of evenly spaced lines going in one direction. If you place an electron in this field, the electron will experience the same force no matter where it is within that region. In our problem, the magnetic field is perpendicular to the motion of the electrons. This position is key because it ensures the magnetic force acts as a centripetal force, causing the electrons to move in a circular path.

Electrons are negatively charged particles. When they move, they create an electric current and respond to magnetic fields. The direction of the force experienced by a moving electron in a magnetic field is given by the right-hand rule:

• Point your right thumb in the direction of the electron's velocity.
• Point your fingers in the direction of the magnetic field.
• Your palm force direction is the direction of the force on a positive charge, but since electrons are negative, the force is in the opposite direction.

In the exercise, the electrons are moving at a speed of \(1.3 \times 10^6 \) meters per second. Because the magnetic field is perpendicular to their motion, the force will be perpendicular to both the velocity of the electrons and the magnetic field lines. This perpendicular force causes the electrons to move in a circular path.

The centripetal force is the force that keeps an object moving in a circular path. It always points towards the center of the circle. For an object of mass \(m\), moving with a velocity \(v\), in a path of radius \(r\), the centripetal force \(F_C\) is given by:

\[ F_C = \frac{m v^2}{r} \]

In our scenario, the magnetic force is acting as the centripetal force. That means we can set the magnetic force equal to the centripetal force:

\[ q v B = \frac{m v^2}{r} \]

where \(q\) is the charge of the electron, \(v\) is the velocity, \(B\) is the magnetic field, and \(r\) is the radius of the circular path. This balance allows us to solve for the magnetic field strength.

Lastly, we can calculate the magnetic field strength. Rearranging the equation given by the balance of forces, we get:

\[ B = \frac{m v}{q r} \]

We know the mass of an electron \(m\) is \(9.11 \times 10^{-31}\) kilograms, its charge \(q\) is \(1.6 \times 10^{-19}\) coulombs, the velocity \(v\) is \(1.3 \times 10^6 \) meters per second, and the radius \(r\) is 0.35 meters. Plugging in these values, the magnetic field strength is:

\[ B = \frac{(9.11 \times 10^{-31}\ \text{kg})(1.3 \times 10^6 \ \text{m/s})}{(1.6 \times 10^{-19} \ \text{C})(0.35 \ \text{m})} \]

Performing the calculation, we get:

\[ B = 2.1 \times 10^{-2} \ \text{T} \]

The magnetic field strength is 2.1 \times 10^{-2}\ tesla, which is what completes our solution. Understanding these relationships and how to manipulate the formulas is crucial for solving problems involving magnetic fields and electron motions.