To calculate the radius of the orbit of an electron moving in a uniform magnetic field, we will use the formula for the radius of the circular path of a charged particle in a magnetic field. The radius [tex]\( r \)[/tex] is given by:
[tex]\[ r = \frac{m \cdot v}{q \cdot B} \][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the electron
- [tex]\( v \)[/tex] is the speed of the electron
- [tex]\( q \)[/tex] is the charge of the electron
- [tex]\( B \)[/tex] is the magnetic field strength
Given:
- Speed of the electron, [tex]\( v = 2.00 \times 10^7 \, \text{m/s} \)[/tex]
- Magnetic field strength, [tex]\( B = 1.20 \times 10^{-3} \, \text{T} \)[/tex]
- Mass of the electron, [tex]\( m = 9.11 \times 10^{-31} \, \text{kg} \)[/tex]
- Charge of the electron, [tex]\( q = 1.61 \times 10^{-19} \, \text{C} \)[/tex]
Now, let's substitute these values into the formula.
[tex]\[ r = \frac{(9.11 \times 10^{-31} \, \text{kg}) \cdot (2.00 \times 10^7 \, \text{m/s})}{(1.61 \times 10^{-19} \, \text{C}) \cdot (1.20 \times 10^{-3} \, \text{T})} \][/tex]
We calculate the numerator first:
[tex]\[ 9.11 \times 10^{-31} \, \text{kg} \times 2.00 \times 10^7 \, \text{m/s} = 1.822 \times 10^{-23} \, \text{kg} \cdot \text{m/s} \][/tex]
Next, we calculate the denominator:
[tex]\[ 1.61 \times 10^{-19} \, \text{C} \times 1.20 \times 10^{-3} \, \text{T} = 1.932 \times 10^{-22} \, \text{C} \cdot \text{T} \][/tex]
Now, we divide the numerator by the denominator to find the radius:
[tex]\[ r = \frac{1.822 \times 10^{-23}}{1.932 \times 10^{-22}} \approx 0.0943 \, \text{m} \][/tex]
Therefore, the radius of the orbit is approximately [tex]\( 0.0943 \)[/tex] meters.
