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A non conducting disc of radius R, having a concentric hole of radius R/2 carries a uniform positive surface charge of area density σ on one face. It is rotated with an angular velocity ω about a perpendicular axis passing through its center. Its magnetic moment is
- a)15πσωR4/64
- b)5πσωR4/32
- c)πσωR4/32
- d)None
Correct answer is option 'A'. Can you explain this answer?
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To find the electric field due to the non-conducting disc, we can use Gauss's Law. Gauss's Law states that the electric flux through a closed surface is proportional to the charge enclosed by the surface.
Let's consider a Gaussian surface in the shape of a cylinder with radius r and height h, centered on the disc. The electric field is constant and perpendicular to the surface of the disc, so the electric flux through the curved surface of the cylinder is zero.
Therefore, the total electric flux through the Gaussian surface is equal to the flux through the top and bottom surfaces of the cylinder.
The flux through each of the top and bottom surfaces is given by:
Φ = E * A
where Φ is the electric flux, E is the electric field, and A is the area of the surface.
The area of each surface is:
A = π * r^2
The charge enclosed by the Gaussian surface is the total charge on the disc, which is equal to the surface charge density multiplied by the area of the disc:
Q = σ * π * R^2
where Q is the charge, σ is the surface charge density, and R is the radius of the disc.
According to Gauss's Law, the electric flux through the Gaussian surface is equal to the charge enclosed divided by the permittivity of free space (ε₀):
Φ = Q / ε₀
Setting the two expressions for the electric flux equal to each other, we get:
E * A = Q / ε₀
Substituting the values for A and Q, we have:
E * π * r^2 = (σ * π * R^2) / ε₀
Simplifying, we find:
E = (σ * R^2) / (2 * ε₀ * r^2)
Therefore, the electric field due to the non-conducting disc at a distance r from the center of the disc is given by:
E = (σ * R^2) / (2 * ε₀ * r^2)
Let's consider a Gaussian surface in the shape of a cylinder with radius r and height h, centered on the disc. The electric field is constant and perpendicular to the surface of the disc, so the electric flux through the curved surface of the cylinder is zero.
Therefore, the total electric flux through the Gaussian surface is equal to the flux through the top and bottom surfaces of the cylinder.
The flux through each of the top and bottom surfaces is given by:
Φ = E * A
where Φ is the electric flux, E is the electric field, and A is the area of the surface.
The area of each surface is:
A = π * r^2
The charge enclosed by the Gaussian surface is the total charge on the disc, which is equal to the surface charge density multiplied by the area of the disc:
Q = σ * π * R^2
where Q is the charge, σ is the surface charge density, and R is the radius of the disc.
According to Gauss's Law, the electric flux through the Gaussian surface is equal to the charge enclosed divided by the permittivity of free space (ε₀):
Φ = Q / ε₀
Setting the two expressions for the electric flux equal to each other, we get:
E * A = Q / ε₀
Substituting the values for A and Q, we have:
E * π * r^2 = (σ * π * R^2) / ε₀
Simplifying, we find:
E = (σ * R^2) / (2 * ε₀ * r^2)
Therefore, the electric field due to the non-conducting disc at a distance r from the center of the disc is given by:
E = (σ * R^2) / (2 * ε₀ * r^2)
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A non conducting disc of radius R, having a concentric hole of radius R/2 carries a uniform positive surface charge of area density σ on one face. It is rotated with an angular velocity ω about a perpendicular axis passing through its center. Its magnetic moment isa)15πσωR4/64b)5πσωR4/32c)πσωR4/32d)NoneCorrect answer is option 'A'. Can you explain this answer?
Question Description
A non conducting disc of radius R, having a concentric hole of radius R/2 carries a uniform positive surface charge of area density σ on one face. It is rotated with an angular velocity ω about a perpendicular axis passing through its center. Its magnetic moment isa)15πσωR4/64b)5πσωR4/32c)πσωR4/32d)NoneCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A non conducting disc of radius R, having a concentric hole of radius R/2 carries a uniform positive surface charge of area density σ on one face. It is rotated with an angular velocity ω about a perpendicular axis passing through its center. Its magnetic moment isa)15πσωR4/64b)5πσωR4/32c)πσωR4/32d)NoneCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A non conducting disc of radius R, having a concentric hole of radius R/2 carries a uniform positive surface charge of area density σ on one face. It is rotated with an angular velocity ω about a perpendicular axis passing through its center. Its magnetic moment isa)15πσωR4/64b)5πσωR4/32c)πσωR4/32d)NoneCorrect answer is option 'A'. Can you explain this answer?.
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