Question

An electron accelerates horizontally from rest through an electric field and into a magnetic field. The charged plates forming the electric field are arranged vertically, with a potential difference of 25 V between them. The magnetic field has a magnitude of 0.50 T, directed into the plane/page. (me = 9.1 x 10⁻³¹ kg and e = 1.6 x 10⁻¹⁹ C.)

Calculate the initial speed of the electron upon entering the magnetic field. [T-1, A-1]

Answer 1

To calculate the initial speed of the electron upon entering the magnetic field, we can use the equation F = qvB, where F is the force experienced by the electron, q is the charge of the electron, v is the initial speed of the electron, and B is the magnitude of the magnetic field.

Step-by-step explanation:

To calculate the initial speed of the electron upon entering the magnetic field, we can use the equation:

F = qvB

Where F is the force experienced by the electron, q is the charge of the electron, v is the initial speed of the electron, and B is the magnitude of the magnetic field.

In this case, the force experienced by the electron is equal to ma, where m is the mass of the electron and a is its acceleration.

We can rearrange the equation to solve for the initial speed:

v = F/(qB)

Substituting the given values, we have:

v = (ma)/(qB)

Now, we can plug in the known values:

v = (9.1x10^-31 kg x a)/(1.6x10^-19 C x 0.50 T)

Given that the potential difference between the plates is 25 V, we can calculate the acceleration using the equation:

a = ΔV/d

where ΔV is the potential difference and d is the distance between the plates.

Since the plates are arranged vertically, the distance between them is not given. Therefore, the problem cannot be solved with the given information.