Question

In a H-atom, an electron moves in a circular orbit of radius $5.2 \times 10^{-11}\( meter and produces a mag. field of \)12.56$ Tesla at its nucleus. The current produced by the motion of the electron will be (a) \(6.53 \times 10^{-3}\) (b) \(13.25 \times 10^{-10}\) (c) \(9.6 \times 10^{6}\) (d) \(1.04 \times 10^{-3}\)

Short answer

The current produced by the motion of the electron in the given hydrogen atom is approximately \(1.04 \times 10^{-3} A\). The correct answer is (d) \(1.04 \times 10^{-3}\).

Step-by-step solution

Write down the relevant formula for magnetic field produced by a current-carrying loop

The magnetic field produced by a current-carrying loop at its center is given by the formula: \(B = \dfrac{\mu_0 I}{2 r}\), where - \(B\) is the magnetic field, - \(\mu_0\) is the magnetic constant (\(4\pi \times 10^{-7} Tm/A\)) - \(I\) is the current, - and \(r\) is the radius of the loop.

Solve for the current (I) using the given values

We are given the magnetic field \(B = 12.56\) T and the radius \(r = 5.2 \times 10^{-11}\) m. Plug these values into the formula and solve for the current: \(I = \dfrac{2 r B}{\mu_0}\). Substituting the values, we have: \(I = \dfrac{2 \times (5.2 \times 10^{-11}) \times 12.56}{(4\pi \times 10^{-7})}\)

Calculate the current I

Calculate the current I: \(I = \dfrac{2 \times (5.2 \times 10^{-11}) \times 12.56}{(4\pi \times 10^{-7})} \approx 1.04 \times 10^{-3} A\) Thus, the current produced by the motion of the electron is approximately \(1.04 \times 10^{-3} A\). The correct answer is (d) \(1.04 \times 10^{-3}\).