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A uniformly charged ring of radius 0.1 m rotates at a frequency of 104 rps about its axis. The ratio of energy density of electric field to the energy density of the magnetic field at a point on the axis at distance 0.2 m from the centre is in form X × 109. Find the value of X. (Use speed of light c = 3 × 108 m/s, π2 = 10)
Correct answer is '9'. Can you explain this answer?
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A uniformly charged ring of radius 0.1 m rotates at a frequency of 104...
Electric field at P is
Magnetic field at P is 
f = frequency of revolution.
Electric energy density
Magnetic energy density 
f = frequency of revolution.
Electric energy density
Magnetic energy density Most Upvoted Answer
A uniformly charged ring of radius 0.1 m rotates at a frequency of 104...
The energy density of the electric field at a point on the axis of a uniformly charged ring can be given by:
u_e = (1/2)ε0E^2
where ε0 is the vacuum permittivity and E is the electric field strength.
The energy density of the magnetic field at the same point can be given by:
u_m = (1/2)B^2/μ0
where μ0 is the vacuum permeability and B is the magnetic field strength.
To find the ratio of energy densities, we need to find the values of E and B at the given point.
The electric field at a point on the axis of a uniformly charged ring can be given by:
E = (kQz) / (2πε0r^3)
where k is the Coulomb constant, Q is the total charge of the ring, z is the distance from the ring's center, and r is the radius of the ring.
The magnetic field at a point on the axis of a uniformly charged ring can be given by:
B = (μ0I) / (2r)
where I is the current of the ring.
Given that the ring has a radius of 0.1m and rotates at a frequency of 104 rps, we can calculate the current of the ring:
I = 2πfQ
where f is the frequency of rotation.
Plugging in the values, we have:
I = 2π(104)Q
Now, we can find the values of E and B at the point on the axis at a distance of 0.2m from the center:
z = 0.2m
r = 0.1m
E = (kQz) / (2πε0r^3)
= (9x10^9)(Q)(0.2) / (2π(8.85x10^-12)(0.1)^3)
B = (μ0I) / (2r)
= (4πx10^-7)(2π(104)Q) / (2(0.1))
Now, we can calculate the ratio of energy densities:
u_e/u_m = [(1/2)ε0E^2] / [(1/2)B^2/μ0]
= (ε0E^2) / (B^2/μ0)
= (ε0E^2μ0) / B^2
= (E^2 / B^2) * (ε0/μ0)
We can substitute the values of E and B into the equation:
u_e/u_m = [(9x10^9)(Q)(0.2) / (2π(8.85x10^-12)(0.1)^3)^2] / [((4πx10^-7)(2π(104)Q) / (2(0.1)))^2] * (8.85x10^-12 / 4πx10^-7)
Simplifying the equation further, we can cancel out common terms:
u_e/u_m = (9x10^9)(0.2) / (2π(8.85x10^-12)(0.1)^3)^2 / ((4πx10^-7)(2π(104))(0.1))^2 * (
u_e = (1/2)ε0E^2
where ε0 is the vacuum permittivity and E is the electric field strength.
The energy density of the magnetic field at the same point can be given by:
u_m = (1/2)B^2/μ0
where μ0 is the vacuum permeability and B is the magnetic field strength.
To find the ratio of energy densities, we need to find the values of E and B at the given point.
The electric field at a point on the axis of a uniformly charged ring can be given by:
E = (kQz) / (2πε0r^3)
where k is the Coulomb constant, Q is the total charge of the ring, z is the distance from the ring's center, and r is the radius of the ring.
The magnetic field at a point on the axis of a uniformly charged ring can be given by:
B = (μ0I) / (2r)
where I is the current of the ring.
Given that the ring has a radius of 0.1m and rotates at a frequency of 104 rps, we can calculate the current of the ring:
I = 2πfQ
where f is the frequency of rotation.
Plugging in the values, we have:
I = 2π(104)Q
Now, we can find the values of E and B at the point on the axis at a distance of 0.2m from the center:
z = 0.2m
r = 0.1m
E = (kQz) / (2πε0r^3)
= (9x10^9)(Q)(0.2) / (2π(8.85x10^-12)(0.1)^3)
B = (μ0I) / (2r)
= (4πx10^-7)(2π(104)Q) / (2(0.1))
Now, we can calculate the ratio of energy densities:
u_e/u_m = [(1/2)ε0E^2] / [(1/2)B^2/μ0]
= (ε0E^2) / (B^2/μ0)
= (ε0E^2μ0) / B^2
= (E^2 / B^2) * (ε0/μ0)
We can substitute the values of E and B into the equation:
u_e/u_m = [(9x10^9)(Q)(0.2) / (2π(8.85x10^-12)(0.1)^3)^2] / [((4πx10^-7)(2π(104)Q) / (2(0.1)))^2] * (8.85x10^-12 / 4πx10^-7)
Simplifying the equation further, we can cancel out common terms:
u_e/u_m = (9x10^9)(0.2) / (2π(8.85x10^-12)(0.1)^3)^2 / ((4πx10^-7)(2π(104))(0.1))^2 * (
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A uniformly charged ring of radius 0.1 m rotates at a frequency of 104rps about its axis. The ratio of energy density of electric field to the energy density of the magnetic field at a point on the axis at distance 0.2 m from the centre is in form X × 109. Find the value of X. (Use speed of light c = 3 × 108 m/s, π2= 10)Correct answer is '9'. Can you explain this answer?
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A uniformly charged ring of radius 0.1 m rotates at a frequency of 104rps about its axis. The ratio of energy density of electric field to the energy density of the magnetic field at a point on the axis at distance 0.2 m from the centre is in form X × 109. Find the value of X. (Use speed of light c = 3 × 108 m/s, π2= 10)Correct answer is '9'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A uniformly charged ring of radius 0.1 m rotates at a frequency of 104rps about its axis. The ratio of energy density of electric field to the energy density of the magnetic field at a point on the axis at distance 0.2 m from the centre is in form X × 109. Find the value of X. (Use speed of light c = 3 × 108 m/s, π2= 10)Correct answer is '9'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniformly charged ring of radius 0.1 m rotates at a frequency of 104rps about its axis. The ratio of energy density of electric field to the energy density of the magnetic field at a point on the axis at distance 0.2 m from the centre is in form X × 109. Find the value of X. (Use speed of light c = 3 × 108 m/s, π2= 10)Correct answer is '9'. Can you explain this answer?.
A uniformly charged ring of radius 0.1 m rotates at a frequency of 104rps about its axis. The ratio of energy density of electric field to the energy density of the magnetic field at a point on the axis at distance 0.2 m from the centre is in form X × 109. Find the value of X. (Use speed of light c = 3 × 108 m/s, π2= 10)Correct answer is '9'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A uniformly charged ring of radius 0.1 m rotates at a frequency of 104rps about its axis. The ratio of energy density of electric field to the energy density of the magnetic field at a point on the axis at distance 0.2 m from the centre is in form X × 109. Find the value of X. (Use speed of light c = 3 × 108 m/s, π2= 10)Correct answer is '9'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniformly charged ring of radius 0.1 m rotates at a frequency of 104rps about its axis. The ratio of energy density of electric field to the energy density of the magnetic field at a point on the axis at distance 0.2 m from the centre is in form X × 109. Find the value of X. (Use speed of light c = 3 × 108 m/s, π2= 10)Correct answer is '9'. Can you explain this answer?.
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