An electron of mass 9.11 × 10-31 kg and charge 1.6 × 10-19 C passes through a cathode ray tube with a velocity of 3.7 × 107 m/s. It enters a magnetic field of flux density 0.47 mT at a right angle. What is the radius of curvature of the path in the magnetic field?
To determine the radius of curvature of the path of the electron in the magnetic field, we can use the equation for the magnetic force on a charged particle moving in a magnetic field.
The magnetic force (F) on a charged particle moving in a magnetic field can be given by the equation:
F = q * v * B * sin(θ)
where:
- F is the magnetic force on the charged particle
- q is the charge of the particle
- v is the velocity of the particle
- B is the magnetic flux density
- θ is the angle between the velocity vector of the particle and the magnetic field vector
In this case, the velocity of the electron (v) is given as 3.7 × 10^7 m/s and the magnetic flux density (B) is given as 0.47 mT (which can be converted to Tesla by dividing by 1000).
Since the angle θ is 90 degrees (as the electron enters the magnetic field at a right angle), the sin(θ) term in the equation becomes 1.
Using all the given values, we can calculate the magnetic force (F) on the electron using the equation above.
F = (1.6 × 10^(-19) C) * (3.7 × 10^7 m/s) * (0.47 × 10^(-3) T) * 1
Simplifying this equation, we get:
F = (1.6 × 3.7 × 0.47 × 10^(-19) × 10^7 × 10^(-3)) C * m/s * T
= 8.928 × 10^(-19) N
The magnetic force (F) acting on the electron can also be expressed as:
F = (m * a) / r
where:
- m is the mass of the electron
- a is the acceleration of the electron
- r is the radius of curvature of the path
Since the electron is moving in a circular path, its acceleration can be calculated using the equation:
a = v^2 / r
Substituting the value of a into the equation for F, we get:
F = (m * v^2) / r
Rearranging the equation to solve for r, we have:
r = (m * v^2) / F
Substituting the values of m, v, and F, we can calculate the radius of curvature (r) of the electron's path in the magnetic field.
r = (9.11 × 10^(-31) kg * (3.7 × 10^7 m/s)^2) / (8.928 × 10^(-19) N)
Simplifying this equation, we get:
r = (9.11 × 3.7^2 × 8.928) / (10^(-31) × 10^(-19))
r = 1.271 × 10^(-4) meters
Therefore, the radius of curvature of the path of the electron in the magnetic field is approximately 1.271 × 10^(-4) meters.
To find the radius of curvature of the path of the electron in the magnetic field, we can use the formula for the magnetic force experienced by a charged particle moving through a magnetic field:
F = q * v * B
where F is the magnetic force, q is the charge of the particle, v is its velocity, and B is the magnetic field flux density.
In this case, the charge of the electron is -1.6 × 10^-19 C, its velocity is 3.7 × 10^7 m/s, and the magnetic field flux density is 0.47 mT (which is equivalent to 0.47 × 10^-3 T). Plugging these values into the formula, we get:
F = (-1.6 × 10^-19 C) * (3.7 × 10^7 m/s) * (0.47 × 10^-3 T)
Simplifying, we find:
F = -8.912 × 10^-5 N
The magnetic force acts as a centripetal force on the electron, causing it to move in a circular path. The centripetal force can be calculated using the formula:
F = (m * v^2) / r
where m is the mass of the particle, v is its velocity, and r is the radius of curvature of the circular path.
In this case, the mass of the electron is 9.11 × 10^-31 kg and the velocity is 3.7 × 10^7 m/s. Plugging these values into the formula, we can solve for the radius of curvature (r):
-8.912 × 10^-5 N = (9.11 × 10^-31 kg) * (3.7 × 10^7 m/s)^2 / r
Simplifying and rearranging the equation, we find:
r = (9.11 × 10^-31 kg) * (3.7 × 10^7 m/s)^2 / -8.912 × 10^-5 N
Plugging in the given values, we can calculate the radius of curvature (r).