Question

An electron accelerated from rest through a voltage of 790 V enters a region of constant magnetic field. If the electron follows a circular path with a radius of 27 cm, what is the magnitude of the magnetic field?

Answer 1

To find the magnitude of the magnetic field, calculate the electron's kinetic energy from the voltage, determine its speed, and then use the centripetal force relationship involving the magnetic field, speed, and radius of the circular path.

Step-by-step explanation:

The question is asking to determine the magnitude of the magnetic field that causes an electron, which was accelerated through a potential of 790 V from rest, to move in a circular path with a radius of 27 cm.

First, we calculate the kinetic energy (KE) gained by the electron using the voltage (V) it was accelerated through: KE = eV, where e is the charge of the electron (1.60 × 10⁻¹⁹ C). Then, using the relationship KE = 0.5mv², where m is the mass of the electron (9.11 × 10⁻³¹ kg), we can find the speed (v) of the electron.

Once the speed is known, the radius (r) of the circular path and the speed (v) can be used to determine the magnitude of the magnetic field (B) using the formula for the centripetal force exerted by the magnetic field: F = evB = mv²/r, solving for B gives us B = mv/(er).