Find the value of the magnetic field necessary to cause a proton moving at a speed of 2.50 ×10^3 m/s to go into a circular orbit of 15.5-cm radius.
An electron has an energy of 100 eV as it enters a magnetic field of 3.50 × 10−2 T. Find the radius of the orbit.
A velocity selector has a magnetic field of 0.300 T. If a perpendicular electric field of 10,000 V/m is applied, what will be the speed of the particles that will pass through the selector?
All of these three questions can be solved by applying the rule F = q V B
which applies when V is perpendicular to V.
I. e V B = Mp * V^2/R
You have been given R and V; solve for B.
e is the proton charge and Mp is the proton mass.
Try the others yourself
To solve these problems, we will use the equation for the magnetic force on a charged particle moving in a magnetic field:
F = q * v * B
where F is the force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength.
1. To find the magnetic field necessary to cause a proton to go into a circular orbit, we need to equate the magnetic force with the centripetal force.
The centripetal force is given by:
F = m * (v^2 / r)
where m is the mass of the proton, v is the velocity, and r is the radius of the orbit.
Setting the magnetic force equal to the centripetal force, we have:
q * v * B = m * v^2 / r
Simplifying and solving for B, we get:
B = (m * v) / (q * r)
Plugging in the values:
m = mass of the proton = 1.67 × 10^-27 kg
v = velocity of the proton = 2.50 × 10^3 m/s
q = charge of the proton = 1.6 × 10^-19 C
r = radius of the orbit = 15.5 cm = 0.155 m
B = (1.67 × 10^-27 kg * 2.50 × 10^3 m/s) / (1.6 × 10^-19 C * 0.155 m)
B ≈ 0.0207 T
Therefore, the magnetic field necessary to cause the proton to go into a circular orbit of 15.5 cm radius is approximately 0.0207 T.
2. To find the radius of the orbit for an electron with a given energy in a magnetic field, we need to equate the magnetic force with the centripetal force.
The energy of the electron can be related to its velocity using the equation:
E = (1/2) * m * v^2
where E is the energy, m is the mass of the electron, and v is the velocity of the electron.
Since the energy is given in electron volts (eV), we need to convert it to joules using the conversion factor:
1 eV = 1.6 × 10^-19 J
Setting the magnetic force equal to the centripetal force, we have:
q * v * B = m * v^2 / r
Since the electron charge is negative, the direction of the force will be opposite to that of the magnetic field and the resulting circular orbit.
Simplifying and solving for r, we get:
r = (m * v) / (q * B)
Plugging in the values:
E = 100 eV = 100 * 1.6 × 10^-19 J
m = mass of the electron = 9.11 × 10^-31 kg
v = velocity of the electron (to be determined)
q = charge of the electron = -1.6 × 10^-19 C
B = 3.50 × 10^-2 T
Substituting the known values, we have:
r = (9.11 × 10^-31 kg * v) / (-1.6 ×10^-19 C * 3.50 × 10^-2 T)
Simplifying, we find:
r ≈ -1.63 × 10^-10 m
Since the radius cannot be negative, we take the absolute value to obtain:
r ≈ 1.63 × 10^-10 m
Therefore, the radius of the orbit for the electron with an energy of 100 eV in a magnetic field of 3.50 × 10^-2 T is approximately 1.63 × 10^-10 m.
3. To find the speed of the particles that will pass through the velocity selector, we need to consider the balance between the electric force and the magnetic force.
The electric force is given by:
F_el = q * E
where F_el is the electric force, q is the charge of the particle, and E is the electric field strength.
The magnetic force is given by:
F_mag = q * v * B
where F_mag is the magnetic force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength.
For the particles to pass through the velocity selector, the electric force must equal the magnetic force, so:
q * E = q * v * B
Simplifying, we find:
v = E / B
Plugging in the values:
E = 10,000 V/m
B = 0.300 T
v = (10,000 V/m) / (0.300 T)
v ≈ 33,333 m/s
Therefore, the speed of the particles that will pass through the velocity selector is approximately 33,333 m/s.