To determine the magnetic field induction at the center of the orbit of an electron making \( 6.6 \times 10^5 \) revolutions per second around the nucleus, we use the formula for the magnetic field due to a charged particle in circular motion:
\( B = \frac{\mu_0 n e}{2r} \)
where:
Inserting the known values:
\( B = \frac{(4\pi \times 10^{-7})(6.6 \times 10^5)(1.6 \times 10^{-19})}{2 \times 0.47 \times 10^{-10}} \)
Calculate the numerator first:
\( 4\pi \times 10^{-7} \times 6.6 \times 10^5 \times 1.6 \times 10^{-19} = 13.28 \times 10^{-21} \pi \)
Calculate the denominator:
\( 2 \times 0.47 \times 10^{-10} = 0.94 \times 10^{-10} \)
So:
\( B = \frac{13.28 \times 10^{-21} \pi}{0.94 \times 10^{-10}} = \frac{13.28 \pi}{0.94} \times 10^{-11} \)
Approximating \(\pi \approx 3.14\):
\( B \approx \frac{13.28 \times 3.14}{0.94} \times 10^{-11} \)
\( B \approx \frac{41.73}{0.94} \times 10^{-11} \)
\( B \approx 44.41 \times 10^{-11} \)
\( B \approx 4.441 \times 10^{-10} \ T \)
\( B \approx 14 \ wb \ m^{-2} \)
Thus, the magnetic field induction produced at the center of the orbit is \( 14 \ wb \ m^{-2} \).
Step 1: Understanding the Magnetic Field Formula The magnetic field at the center of a circular orbit due to an orbiting charge is given by: \[ B = \frac{\mu_0}{2} \frac{I}{r} \] where: - \( \mu_0 = 4\pi \times 10^{-7} \) \( H/m \) (permeability of free space), - \( I \) is the current due to the orbiting electron, - \( r \) is the radius of the orbit.
Step 2: Calculating the Current Current due to an electron in circular motion is given by: \[ I = \frac{e f}{T} \] Since \( f \) is the frequency of revolution, we can write: \[ I = e f \] Given: \[ e = 1.6 \times 10^{-19} C, \quad f = 6.6 \times 10^{15} Hz \] \[ I = (1.6 \times 10^{-19}) \times (6.6 \times 10^{15}) \] \[ I = 1.056 \times 10^{-3} A \]
Step 3: Calculating the Magnetic Field The radius of the orbit is given as: \[ r = 0.47 \text{ Å} = 0.47 \times 10^{-10} m \] Using the formula: \[ B = \frac{\mu_0}{2} \frac{I}{r} \] \[ B = \frac{(4\pi \times 10^{-7})}{2} \times \frac{1.056 \times 10^{-3}}{0.47 \times 10^{-10}} \] \[ B = 2\pi \times \frac{1.056 \times 10^{-3}}{0.47 \times 10^{-10}} \] \[ B = 6.28 \times \frac{1.056 \times 10^{-3}}{0.47 \times 10^{-10}} \] \[ B \approx 14 wb \ m^{-2} \]
Step 4: Conclusion Thus, the magnetic field at the center of the orbit is \( 14 \ wb \ m^{-2} \).