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An atom consists of a Proton and electron separated by a distance of 0.5 A. assuming that the electron moves in a circular orbit with a velocity and 10¹³ cm / sec find the magnetic flux density at the region of nucleus?
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An atom consists of a Proton and electron separated by a distance of 0...
It is not possible to calculate the magnetic flux density at the region of nucleus without knowing the charge and magnetic moment of the proton and the electron, as well as the radius of the electron's orbit.
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An atom consists of a Proton and electron separated by a distance of 0...
Magnetic Flux Density Calculation
To find the magnetic flux density (B) at the region of the nucleus due to an electron moving in a circular orbit, we can use the following steps:
1. Parameters Given
- **Distance (r)**: 0.5 Å = \(0.5 \times 10^{-10} \, m\)
- **Velocity (v)**: \(10^{13} \, cm/s = 10^{11} \, m/s\)
2. Current Equivalent of Electron Motion
The electron moving in a circular path can be treated as a current loop. The equivalent current (I) due to the electron can be calculated as:
- **Charge of electron (e)**: \(1.6 \times 10^{-19} \, C\)
- **Frequency (f)**: The frequency of revolution can be calculated using the velocity and radius:
\[
f = \frac{v}{2\pi r}
\]
Substituting the values:
\[
f = \frac{10^{11}}{2\pi \times 0.5 \times 10^{-10}} \approx 3.18 \times 10^{20} \, Hz
\]
- **Current (I)**:
\[
I = e \times f \approx 1.6 \times 10^{-19} \times 3.18 \times 10^{20} \approx 5.09 \, A
\]
3. Magnetic Flux Density (B)
The magnetic flux density at the center of the circular loop can be calculated using the formula:
\[
B = \frac{\mu_0 I}{2r}
\]
Where:
- **\(\mu_0\)** (permeability of free space) = \(4\pi \times 10^{-7} \, T \cdot m/A\)
Substituting the values:
\[
B = \frac{4\pi \times 10^{-7} \times 5.09}{2 \times 0.5 \times 10^{-10}}
\]
Calculating this gives:
\[
B \approx 1.3 \times 10^{3} \, T
\]
Conclusion
The magnetic flux density at the region of the nucleus due to the electron's motion is approximately \(1.3 \, kT\). This strong magnetic field is a result of the high velocity and the small radius of the electron’s orbit.
To find the magnetic flux density (B) at the region of the nucleus due to an electron moving in a circular orbit, we can use the following steps:
1. Parameters Given
- **Distance (r)**: 0.5 Å = \(0.5 \times 10^{-10} \, m\)
- **Velocity (v)**: \(10^{13} \, cm/s = 10^{11} \, m/s\)
2. Current Equivalent of Electron Motion
The electron moving in a circular path can be treated as a current loop. The equivalent current (I) due to the electron can be calculated as:
- **Charge of electron (e)**: \(1.6 \times 10^{-19} \, C\)
- **Frequency (f)**: The frequency of revolution can be calculated using the velocity and radius:
\[
f = \frac{v}{2\pi r}
\]
Substituting the values:
\[
f = \frac{10^{11}}{2\pi \times 0.5 \times 10^{-10}} \approx 3.18 \times 10^{20} \, Hz
\]
- **Current (I)**:
\[
I = e \times f \approx 1.6 \times 10^{-19} \times 3.18 \times 10^{20} \approx 5.09 \, A
\]
3. Magnetic Flux Density (B)
The magnetic flux density at the center of the circular loop can be calculated using the formula:
\[
B = \frac{\mu_0 I}{2r}
\]
Where:
- **\(\mu_0\)** (permeability of free space) = \(4\pi \times 10^{-7} \, T \cdot m/A\)
Substituting the values:
\[
B = \frac{4\pi \times 10^{-7} \times 5.09}{2 \times 0.5 \times 10^{-10}}
\]
Calculating this gives:
\[
B \approx 1.3 \times 10^{3} \, T
\]
Conclusion
The magnetic flux density at the region of the nucleus due to the electron's motion is approximately \(1.3 \, kT\). This strong magnetic field is a result of the high velocity and the small radius of the electron’s orbit.
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An atom consists of a Proton and electron separated by a distance of 0.5 A. assuming that the electron moves in a circular orbit with a velocity and 10¹³ cm / sec find the magnetic flux density at the region of nucleus?
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An atom consists of a Proton and electron separated by a distance of 0.5 A. assuming that the electron moves in a circular orbit with a velocity and 10¹³ cm / sec find the magnetic flux density at the region of nucleus? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An atom consists of a Proton and electron separated by a distance of 0.5 A. assuming that the electron moves in a circular orbit with a velocity and 10¹³ cm / sec find the magnetic flux density at the region of nucleus? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An atom consists of a Proton and electron separated by a distance of 0.5 A. assuming that the electron moves in a circular orbit with a velocity and 10¹³ cm / sec find the magnetic flux density at the region of nucleus?.
An atom consists of a Proton and electron separated by a distance of 0.5 A. assuming that the electron moves in a circular orbit with a velocity and 10¹³ cm / sec find the magnetic flux density at the region of nucleus? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An atom consists of a Proton and electron separated by a distance of 0.5 A. assuming that the electron moves in a circular orbit with a velocity and 10¹³ cm / sec find the magnetic flux density at the region of nucleus? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An atom consists of a Proton and electron separated by a distance of 0.5 A. assuming that the electron moves in a circular orbit with a velocity and 10¹³ cm / sec find the magnetic flux density at the region of nucleus?.
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