# 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).