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Chapter 13: Problem 1883

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

Expert verified
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

01

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.
02

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: \(I = \dfrac{2 \times (5.2 \times 10^{-11}) \times 12.56}{(4\pi \times 10^{-7})}\)
03

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) \(1.04 \times 10^{-3}\).

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Most popular questions from this chapter

Force between two identical bar magnets whose centers are I meter apart is \(4.8 \mathrm{~N}\), when their axes are in the same line. If separation is increased to \(2 r\), the force between them is reduced to (a) \(2.4 \mathrm{~N}\) (b) \(1.2 \mathrm{~N}\) (c) \(0.6 \mathrm{~N}\) (d) \(0.3 \mathrm{~N}\)

\(\mathrm{A}\) bar magnet of length \(10 \mathrm{~cm}\) and having the pole strength equal \(10^{3} \mathrm{Am}\) to is kept in a magnetic field having magnetic induction (B) equal to \(4 \pi \times 10^{3}\) tesla. It makes an angle of \(30^{\circ}\) with the direction of magnetic induction. The value of the torque acting on the magnet is Joule. (a) \(2 \pi \times 10^{-7}\) (b) \(2 \pi \times 10^{5}\) (c) \(0.5\) (d) \(0.5 \times 10^{2}\)

The angles of dip at two places are \(30^{\circ}\) and \(45^{\circ}\). The ratio of horizontal components of earth's magnetic field at the two places will be (a) \(\sqrt{3}: \sqrt{2}\) (b) \(1: \sqrt{2}\) (c) \(1: 2\) (d) \(1: \sqrt{3}\)

A magnet of magnetic moment \(\mathrm{M}\) and pole strength \(\mathrm{m}\) is divided in two equal parts, then magnetic moment of each part will be (a) \(\mathrm{M}\) (b) \((\mathrm{M} / 2)\) (c) \((\mathrm{M} / 4)\) (d) \(2 \mathrm{M}\)

Two particles \(\mathrm{X}\) and \(\mathrm{Y}\) having equal charges, after being accelerated through the same potential difference, enter a region of uniform mag. field and describe circular path of radius \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) respectively. The ratio of mass of \(\mathrm{X}\) to that of \(\mathrm{Y}\) is (a) \(\sqrt{\left(r_{1} / \mathrm{r}_{2}\right)}\) (b) \(\left(\mathrm{r}_{2} / \mathrm{r}_{1}\right)\) (c) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)^{2}\) (b) \(\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)\)
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