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A short magnetic moment of 50 A m square Calculate the magnetic field intensity at a distance of 0.2 m from its centre on 1 it's axial line 2 it's equatorial line?
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A short magnetic moment of 50 A m square Calculate the magnetic field ...
For axial position,
B=(u0/4π)*2m/r^3

For equatorial position,
B=(u0/4π)*m/r^3

where r is the distance at which the magnetic field intensity is to be found
Community Answer
A short magnetic moment of 50 A m square Calculate the magnetic field ...
Magnetic Field Intensity at a Distance from a Magnetic Moment

In order to calculate the magnetic field intensity at a certain distance from a magnetic moment, we can use the formula:

B = (μ₀ / 4π) * (3(m⋅r) * r - m * r^2) / r^5

Where:
- B is the magnetic field intensity
- μ₀ is the permeability of free space (μ₀ = 4π × 10^-7 T⋅m/A)
- m is the magnetic moment
- r is the distance from the center of the magnetic moment

Now, let's calculate the magnetic field intensity at a distance of 0.2 m from the center of the magnetic moment.

1. Axial Line

When the distance is measured along the axial line, the value of r will be equal to the distance itself.

Using the given values:
- m = 50 A m²
- r = 0.2 m

Plugging these values into the formula, we get:

B = (μ₀ / 4π) * (3(m⋅r) * r - m * r²) / r^5
B = (4π × 10^-7 T⋅m/A / 4π) * (3(50 A m² ⋅ 0.2 m) * 0.2 m - 50 A m² * (0.2 m)²) / (0.2 m)^5

Simplifying the expression, we find:

B = (10^-7 T⋅m/A) * (0.3 - 0.2) / (0.2 m)^3
B = (10^-7 T⋅m/A) * 0.1 / (0.008 m³)
B = 0.1 T / (8 × 10^-6 m³)
B = 1.25 × 10^4 T/m

Therefore, the magnetic field intensity at a distance of 0.2 m from the center of the magnetic moment along its axial line is 1.25 × 10^4 T/m.

2. Equatorial Line

When the distance is measured along the equatorial line, the value of r will be the perpendicular distance from the center of the magnetic moment.

Using the given values:
- m = 50 A m²
- r = 0.2 m

Plugging these values into the formula, we get:

B = (μ₀ / 4π) * (3(m⋅r) * r - m * r²) / r^5
B = (4π × 10^-7 T⋅m/A / 4π) * (3(50 A m² ⋅ 0.2 m) * 0 - 50 A m² * (0.2 m)²) / (0.2 m)^5

Simplifying the expression, we find:

B = (10^-7 T⋅m/A) * (-0.2) / (0.008 m³)
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The orbital and spin angular momentum of the atom influence its magnetic structure and these properties are most directly studied by placing the atom in a magnetic field. Also, a magnetic field can affect the wavelengths of the emitted photons.The angular momentum vector associated with an atomic state can take up only certain specified directions in space. This concept of space quantization was shown by Otto Stern and Walthor Gerlach in their experiment.In the experiment, silver is vapourized in an electric oven and silver atoms spray into the evacuated apparatus through a small hole in the oven wall. The atoms which are electrically neutral but have a magnetic moment, are formed into a narrow beam as they pass through a slit in a screen. The beam, thus collimated, then passes between the poles of an electromagnet and finally, deposits its silver atoms on a glass plate that serves as a detector. The pole faces of the magnet are shaped to make the magnetic field as nonuniform as possible.In a non-uniform magnetic field, there is a net force on a magnetic dipole. Its magnitude and direction depends on the orientation of the dipole. Thus the silver atoms in the beam are deflected up or down, depending on the orientation of their magnetic dipole moments with respect to the z–direction.The potential energy of a magnetic dipole in a magnetic field where is magnetic dipole moment of the atom. From symmetry, the magnetic field at the beam position has no x or y components i.e.The net force Fz on the dipole isThus, the net force depends, not on the magnitude of the field itself, but on its spatial derivative or gradient.The ResultsIf space quantization did not exist, then could take on any value from + to –, the result would be a spreading out of the beam when the magnet was turned ON. However, the beam was split cleanly into two subbeams, each subbeam corresponding to one of the two permitted orientations of the magnetic moment ofthe silver atom, as shown.In a silver atom, all the spin and orbital magnetic moments of the electrons cancel, except for those of the atoms single valance electron. For this electron the orbital magnetic moment is zero because orbital angular momentum is zero (because for electrons of s–orbit, L = 0), leaving only the spin magnetic moment. This can take up only two orientations in a magnetic field, corresponding to ms = +1/2 and ms = – 1/2. Hence there are two subbeams – and not some other number.Q.A hydrogen atom in ground state passes through a magnetic field that has a gradient of 16mT/m in the vertical direction. If vertical component magnetic moment of the atom is 9.3 × 10–24 J/T, then force on it due to the magnetic moment of the electron is

The orbital and spin angular momentum of the atom influence its magnetic structure and these properties are most directly studied by placing the atom in a magnetic field. Also, a magnetic field can affect the wavelengths of the emitted photons.The angular momentum vector associated with an atomic state can take up only certain specified directions in space. This concept of space quantization was shown by Otto Stern and Walthor Gerlach in their experiment.In the experiment, silver is vapourized in an electric oven and silver atoms spray into the evacuated apparatus through a small hole in the oven wall. The atoms which are electrically neutral but have a magnetic moment, are formed into a narrow beam as they pass through a slit in a screen. The beam, thus collimated, then passes between the poles of an electromagnet and finally, deposits its silver atoms on a glass plate that serves as a detector. The pole faces of the magnet are shaped to make the magnetic field as nonuniform as possible.In a non-uniform magnetic field, there is a net force on a magnetic dipole. Its magnitude and direction depends on the orientation of the dipole. Thus the silver atoms in the beam are deflected up or down, depending on the orientation of their magnetic dipole moments with respect to the z–direction.The potential energy of a magnetic dipole in a magnetic field where is magnetic dipole moment of the atom. From symmetry, the magnetic field at the beam position has no x or y components i.e.The net force Fz on the dipole isThus, the net force depends, not on the magnitude of the field itself, but on its spatial derivative or gradient.The ResultsIf space quantization did not exist, then could take on any value from + to –, the result would be a spreading out of the beam when the magnet was turned ON. However, the beam was split cleanly into two subbeams, each subbeam corresponding to one of the two permitted orientations of the magnetic moment ofthe silver atom, as shown.In a silver atom, all the spin and orbital magnetic moments of the electrons cancel, except for those of the atoms single valance electron. For this electron the orbital magnetic moment is zero because orbital angular momentum is zero (because for electrons of s–orbit, L = 0), leaving only the spin magnetic moment. This can take up only two orientations in a magnetic field, corresponding to ms = +1/2 and ms = – 1/2. Hence there are two subbeams – and not some other number.Q.A hydrogen atom in ground state passes through a magnetic field that has a gradient of 16mT/m in the vertical direction. If vertical component magnetic moment of the atom is 9.3 × 10–24 J/T, then force on it due to the magnetic moment of the electron is

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A short magnetic moment of 50 A m square Calculate the magnetic field intensity at a distance of 0.2 m from its centre on 1 it's axial line 2 it's equatorial line?
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