An electron in a vacuum is first accelerated
by a voltage of 15600 V and then enters a
region in which there is a uniform magnetic
field of 0.689 T at right angles to the direction
of the electron’s motion.
What is the force on the electron due to the
magnetic field?
Answer in units of N.
To find the force on the electron due to the magnetic field, we can use the formula for the magnetic force on a charged particle:
F = q * v * B
where F is the force, q is the charge of the particle, v is its velocity, and B is the magnetic field strength.
In this case, the charge of an electron is q = -1.6 x 10^-19 C (negative because electrons have a negative charge).
Given:
Voltage = 15600 V
Magnetic field strength = 0.689 T
Since the electron is accelerated by the voltage, it gains kinetic energy. The kinetic energy gained by the electron is equal to the work done on it by the electric field.
The equation for the kinetic energy is:
K.E. = q * V
where K.E. is the kinetic energy gained, q is the charge of the electron, and V is the voltage.
Substituting the values:
K.E. = -1.6 x 10^-19 C * 15600 V
K.E. ≈ -2.496 x 10^-15 J
Since the kinetic energy gained by the electron is equal to the work done on it, we can use this energy to find the velocity of the electron using the equation:
K.E. = (1/2) * m * v^2
where m is the mass of the electron and v is its velocity.
Rearranging the equation to solve for v:
v = sqrt((2 * K.E.) / m)
The mass of an electron is approximately 9.11 x 10^-31 kg.
Substituting the values:
v = sqrt((2 * -2.496 x 10^-15 J) / 9.11 x 10^-31 kg)
v ≈ 1.268 x 10^6 m/s
Now, we can calculate the force on the electron due to the magnetic field using the formula:
F = q * v * B
Substituting the values:
F = -1.6 x 10^-19 C * 1.268 x 10^6 m/s * 0.689 T
F ≈ -1.411 x 10^-12 N
Therefore, the force on the electron due to the magnetic field is approximately -1.411 x 10^-12 N.
To find the force on the electron due to the magnetic field, we can use the formula for the magnetic force experienced by a charged particle moving through a magnetic field.
The formula for the magnetic force is given by:
F = q * v * B * sin(theta)
where:
F is the force on the charged particle,
q is the charge of the particle (in this case, the charge of an electron is -1.6 x 10^(-19) C),
v is the velocity of the particle,
B is the magnetic field strength, and
theta is the angle between the velocity vector and the magnetic field vector.
In this case, the electron is moving at right angles to the direction of the magnetic field, so the angle theta is 90 degrees.
We need to calculate the velocity of the electron first. The acceleration of the electron can be found using the voltage and the charge of the electron. The formula for the acceleration is:
a = (q * V) / m
where:
a is the acceleration of the electron,
q is the charge of the electron,
V is the voltage applied, and
m is the mass of the electron (9.11 x 10^(-31) kg).
Plugging in the given values, we can calculate the acceleration of the electron:
a = (-1.6 x 10^(-19) C * 15600 V) / (9.11 x 10^(-31) kg)
Next, we can calculate the velocity of the electron using the formula for constant acceleration:
v^2 = u^2 + 2 * a * s
where:
v is the final velocity,
u is the initial velocity (which we can assume to be zero since the electron starts from rest),
a is the acceleration calculated earlier, and
s is the distance traveled.
Since the electron is moving in a vacuum, there is no friction, so we can assume that the distance traveled is equal to the acceleration distance. Therefore, we can rewrite the equation as:
v^2 = 2 * a * s
Finally, we can find the velocity of the electron by calculating the square root of the expression:
v = sqrt(2 * a * s)
Now that we have the velocity of the electron, we can substitute all the values into the magnetic force formula:
F = q * v * B * sin(theta)
Plugging in the known values, we can calculate the force on the electron.