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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

The direction of mag. field lines close to a straight conductor carrying current will be (a) Along the length of the conductor (b) Radially outward (c) Circular in a plane perpendicular to the conductor (d) Helical

Graph of force per unit length between two long parallel current carrying conductors and the distance between them (a) Straight line (b) Parabola (c) Ellipse (d) Rectangular hyperbola

An electron having mass \(9 \times 10^{-31} \mathrm{~kg}\), charge $1.6 \times 10^{-19} \mathrm{C}\( and moving with a velocity of \)10^{6} \mathrm{~m} / \mathrm{s}$ enters a region where mag. field exists. If it describes a circle of radius \(0.10 \mathrm{~m}\), the intensity of magnetic field must be Tesla (a) \(1.8 \times 10^{-4}\) (b) \(5.6 \times \overline{10^{-5}}\) (c) \(14.4 \times 10^{-5}\) (d) \(1.3 \times 10^{-6}\)

The unit of ele. current "AMPERE" is the current which when flowing through each of two parallel wires spaced 1 meter apart in vacuum and of infinite length will give rise to a force between them equal to $\mathrm{N} / \mathrm{m}$ (a) 1 (b) \(2 \times 10^{-7}\) (c) \(1 \times 10^{-2}\) (d) \(4 \pi \times 10^{-7}\)

Two parallel long wires \(\mathrm{A}\) and B carry currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\). \(\left(\mathrm{I}_{2}<\mathrm{I}_{1}\right)\) when \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\) are in the same direction the mag. field at a point mid way between the wires is \(10 \mu \mathrm{T}\). If \(\mathrm{I}_{2}\) is reversed, the field becomes \(30 \mu \mathrm{T}\). The ratio \(\left(\mathrm{I}_{1} / \mathrm{I}_{2}\right)\) is (a) 1 (b) 2 (c) 3 (d) 4
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