Question

In the hydrogen atom, the electron is making \(6.6\times 10^{15}\,rps\). If the radius of orbit is \(0.53\times 10^{-10}\,m\), then magnetic field produced at the centre of the orbit is

• A. 140 T
• B. 12.5 T
• C. 1.4 T
• D. 0.14 T

Hint:

Revolving electron gives current \(I=ef\). Magnetic field at centre of orbit: \(B=\dfrac{\mu_0 I}{2r}\).

Answer 1

The Correct Option is B

Step 1: Current due to revolving electron.
Electron revolving in orbit forms a current:
\[ I = ef \] where \(f\) is revolutions per second.
Given \(f = 6.6\times 10^{15}\,rps\).
\[ I = (1.6\times 10^{-19})(6.6\times 10^{15}) = 1.056\times 10^{-3}\,A \] Step 2: Magnetic field at centre of circular loop.
\[ B = \frac{\mu_0 I}{2r} \] Given \(r = 0.53\times 10^{-10}\,m\).
Step 3: Substitute values.
\[ B = \frac{(4\pi\times 10^{-7})(1.056\times 10^{-3})}{2(0.53\times 10^{-10})} \] \[ B = \frac{4\pi\times 1.056\times 10^{-10}}{1.06\times 10^{-10}} \approx 4\pi \approx 12.5\,T \] Final Answer: \[ \boxed{12.5\ T} \]