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Magnetic Moment Of Revolving Electron And Bohr Magneton MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • 16 Questions around this concept.

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Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency of f Hz. The magnitude of magnetic induction at the centre of the ring is:

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A non-conducting disc of radius R charged uniformly over one side with surface charge density σ rotate about its geometrical axis with an angular velocity. The magnetic moment of the disc is
 

 

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An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the center of the circle. The radius of the circle is proportional to

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A 10 eV electron is circulating in a plane at right angles to a uniform field at magnetic induction $10^{-4} \mathrm{~Wb} / \mathrm{m}^2(=1.0$ gauss $)$. The orbital radius of the electron is.

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An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude:

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Two circular current-carrying coils of radii 3cm and 6cm are each equivalent to a magnetic dipole having equal magnetic moments. The currents through the coils are in the ratio of:

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Two particles each of mass m and charge q, are attached to the two ends of a light right rod of length 2l. The rod is rotated at a constant angular speed about a perpendicular axis passing through its center. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the center of the rod is

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If a magnetic dipole of dipole moment M is rotated through an angle $45^{\circ}$ with respect to the erection of the field H, then the work done is

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A charged particle (charge $q$ ) is moving in a circle of radius R with uniform speed v . The associated magnetic moment $\mu$ is giver by

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Concepts Covered - 1

Magnetic dipole moment of a revolving electron

Magnetic dipole moment of a revolving electron-

                                                     

Let us consider an electron that is revolving around in a circle of radius r with a velocity v. The charge of the electron is e and its mass is m, both of which are constant. The time period T of the electrons’ orbit is - 

                                                                         T=\frac{\text { Circumference }}{\text {Velocity}}=\frac{2 \pi r}{v}

So the current due to motion of electron is -

                                                                                   i = \frac{q}{T} = \frac{-e}{\frac{2 \pi r}{v}} = \frac{-ev}{2\pi r}

Now, as we know that the direction of current is opposite to the direction of motion of electron. Now the magnetic moment is defined as - 

                                                                             \begin{array}{l}{\mu=i A} \\ \\ {\text {So the Magnetic moment of an electron: }} \\ \\ {\mu=\frac{-e v}{2 \pi r} A=\frac{-e v}{2 \pi r} \pi r^{2}} \\ \\ {\mu=\frac{-e r v}{2}}\end{array}

If we divide and multiply by the mass of the electron,

                                                                            \mu=\frac{-e}{2 m_{e}} m_{e} v r

As we have studied that that the angular momentum L is given by:

 

                                                                                        L = mvr

So the above equation can be written as - 

                                                                                      \mu=\frac{-e}{2 m_{e}} L

The negative sign shows that the velocity and current are on opposite directions as shown in the figure given above. Also in the vector form it is written as -

                                                                                       \vec{\mu}=\frac{-e}{2 m_{e}} \vec{L} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)

Now, by Niels Bohr (we have studied this particular concept in Atomic structure of Chemistry and we will study this in detail in the chapter Atoms and Nuclei of Physics), Angular momentum of the electron is given as - 

                                                                              L=n \frac{h}{2 \pi}, n=0, \pm 1, \pm 2 \ldots

Where n is the orbit quantum number and h is the Planck’s constant,

Now by using the equation (1)

                                                                                             \begin{array}{l}{\mu=n \frac{-e}{2 m_{e}} \frac{h}{2 \pi}} \\ \\ {\mu=-n \frac{e h}{4 \pi m_{e}}}\end{array}

If we put n =1, then the equation become - 

                                         \begin{array}{l}{\mu_B=- \frac{e h}{4 \pi m_{e}}}\end{array} = 9.27 \times 10^{-27} J/T                (This is called Bohr Magneton μB)

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Magnetic dipole moment of a revolving electron

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