A beam of electrons enters a pair of croseed electric and magnetic fields in which the electric field strenght of 3.0*104 V m^-1 and magnetic flux density of 1*10^-2 T. If the beam is not deflected from its path by the fields, what must be the speed of the electrons?
Thanks!
e V B must equal e E
That means V = E/B = 3*10^4/1*10^-2
The answer will be in m/s for the SI units you are using for E and B
To determine the speed of the electrons, we can use the formula for the force experienced by a charged particle moving in a uniform magnetic field.
The force on a charged particle moving perpendicular to a magnetic field is given by:
F = B * q * v, where F is the force, B is the magnetic flux density, q is the charge of the particle, and v is the velocity of the particle.
In this case, the force due to the magnetic field is balanced by the force due to the electric field, so we can write:
E = F / q, where E is the electric field strength.
We know that E = 3.0 * 10^4 V/m and B = 1 * 10^(-2) T. Let's assume that the charge of the electrons is -e, where e is the elementary charge.
So, the force due to the magnetic field is F = B * q * v = -B * e * v.
The electric force is given by F = E * q = E * (-e).
Since the two forces are equal, we can set them equal to each other.
-E * e = -B * e * v.
Simplifying the equation, we get:
v = E / B.
Substituting the given values, we get:
v = (3.0 * 10^4 V/m) / (1 * 10^(-2) T).
Calculating the value, we get:
v = 3.0 * 10^6 m/s.
Therefore, the speed of the electrons must be 3.0 * 10^6 m/s.
To solve this problem, we can use the fact that in order for the beam of electrons to go undeflected, the electric force and the magnetic force acting on the electrons must be equal in magnitude.
The electric force (F_e) acting on a charged particle can be calculated using the equation:
F_e = q * E
where q is the charge of the particle (in this case, the charge of an electron) and E is the electric field strength.
The magnetic force (F_B) acting on a charged particle moving through a magnetic field can be calculated using the equation:
F_B = q * v * B
where q is the charge of the particle, v is the velocity of the particle, and B is the magnetic flux density.
Since the beam of electrons is not deflected, we can set F_e equal to F_B:
q * E = q * v * B
The charge of an electron (q) is a constant, so we can cancel it out from both sides of the equation:
E = v * B
We can rearrange the equation to solve for the velocity (v):
v = E / B
Substituting the values given in the problem, we have:
v = (3.0 * 10^4 V m^-1) / (1 * 10^-2 T)
Simplifying the equation, we get:
v = 3.0 * 10^6 m s^-1
Therefore, the speed of the electrons must be 3.0 * 10^6 m s^-1 for the beam to remain undeflected in the given electric and magnetic fields.