The electrons in a beam are moving at 2.8 ×
108 m/s in an electric field of 17000 N/C.
What value must the magnetic field have if
the electrons pass through the crossed fields
undeflected?
Answer in units of µT
q v B = q E, therefore
B = E/v
If E is in N/C and v is in m/s, the B field will be in Teslas
To determine the value of the magnetic field needed for the electrons to pass through the crossed fields undeflected, we can use the following formula:
F_mag = F_elec
Where:
F_mag is the magnetic force acting on the electrons
F_elec is the electric force acting on the electrons
The magnetic force (F_mag) on a charged particle moving in a magnetic field is given by:
F_mag = q * v * B
Where:
q is the charge of the electron (-1.6 × 10^-19 C)
v is the velocity of the electrons (2.8 × 10^8 m/s)
B is the magnetic field strength (unknown)
The electric force (F_elec) on the electrons in an electric field is given by:
F_elec = q * E
Where:
E is the electric field strength (17000 N/C)
Setting F_mag equal to F_elec, we can solve for B:
q * v * B = q * E
Simplifying the equation by canceling out the charge (q):
v * B = E
To solve for B, rearrange the equation:
B = E / v
Plugging in the values:
B = 17000 N/C / (2.8 × 10^8 m/s)
Calculating the result:
B ≈ 0.0607 µT
Therefore, the magnetic field must have a value of approximately 0.0607 µT for the electrons to pass through the crossed fields undeflected.
To determine the value of the magnetic field required for the electrons to pass through the crossed fields undeflected, we can use the concept of the Lorentz force. The Lorentz force is the force experienced by a charged particle moving in both an electric field and a magnetic field.
The Lorentz force is given by the equation:
F = q(E + v × B)
where F is the force experienced by the particle, q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field.
In this case, since the electrons are passing through the crossed fields undeflected, the net force experienced by the electrons must be zero. Therefore, we can equate the Lorentz force to zero:
q(E + v × B) = 0
Given that q is the charge of an electron (1.6 × 10^-19 C), E is the electric field (17000 N/C), and v is the velocity of the electrons (2.8 × 10^8 m/s), we can solve for B.
First, rearrange the equation to solve for B:
v × B = -E
Since the cross product of velocity and magnetic field produces a perpendicular vector, we are looking for the magnitude of the magnetic field. Therefore, take the magnitudes of both sides of the equation:
|v × B| = |-E|
|v| |B| = |E|
|B| = |E| / |v|
Substitute the given values:
|B| = (17000 N/C) / (2.8 × 10^8 m/s)
Simplifying the expression gives:
|B| = 6.07 × 10^-5 T
To convert from Tesla (T) to microtesla (µT), we multiply by 10^6:
|B| = 6.07 × 10^1 µT
Therefore, the value of the magnetic field must be 60.7 µT in order for the electrons to pass through the crossed fields undeflected.