What uniform magnetic field applied perpendicular to a beam of electrons moving at 1.3 x 10⁶ m/s, is required to make the electrons travels in a circular arc of radius 0.35 m?

The correct option is 2.1 x 10^-5 T.

Understanding the Physical Principles:
When a charged particle, such as an electron, moves through a uniform magnetic field, it experiences a magnetic force. If the magnetic field is perpendicular to the velocity of the electron, this force is always perpendicular to both the velocity and the magnetic field. This perpendicular force acts as a centripetal force, causing the electron to move in a circular path. We can determine the required magnetic field strength by equating the magnetic force to the centripetal force.

1. Formulating the Equations:
The magnitude of the magnetic force (

$F\_B$

) acting on a particle with charge

$q$

moving at velocity

$v$

perpendicular to a magnetic field

$B$

is given by:

$$F_B=qvB$$

The centripetal force (

$F\_c$

) required to keep a particle of mass

$m$

moving in a circle of radius

$r$

at speed

$v$

is:

$$F_c=\frac{m{v}^2}{r}$$

2. Deriving the Expression for the Magnetic Field:
By equating the magnetic force to the centripetal force, we get:

$$qvB=\frac{m{v}^2}{r}$$

We can simplify this equation by dividing both sides by

$v$

:

$$qB=\frac{mv}{r}$$

Now, we solve for the magnetic field strength

$B$

:

$$B=\frac{mv}{qr}$$

3. Identifying the Constants and Given Values:
We are given the following values in the problem:
- Speed of the electron,

$v$

= 1.3 x 10⁶ m/s
- Radius of the circular path,

$r$

= 0.35 m
Additionally, we use the standard physical constants for an electron:
- Mass of an electron,

$m$

≈ 9.11 x 10^-31 kg
- Magnitude of the charge of an electron,

$q$

≈ 1.60 x 10^-19 C

4. Step-by-Step Calculation:
Substitute these values into our derived formula for

$B$

:

$$B=\frac{(9.11\times {10}^{-31}kg)\times (1.3\times {10}^{6}m/s)}{(1.60\times {10}^{-19}C)\times (0.35m)}$$

First, calculate the numerator:

$$9.11\times 1.3\times {10}^{-31+6}=11.843\times {10}^{-25}kg·m/s$$

Next, calculate the denominator:

$$1.60\times 0.35\times {10}^{-19}=0.56\times {10}^{-19}C·m$$

Now divide the numerator by the denominator to find the magnetic field:

$$B=\frac{11.843\times {10}^{-25}}{0.56\times {10}^{-19}}T$$

$$B\approx 21.148\times {10}^{-6}T$$

$$B\approx 2.1\times {10}^{-5}T$$

Thus, the required uniform magnetic field strength is 2.1 x 10^-5 T.