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An infinitely long straight wire is carrying a steady current I. The ratio of magnetic energy density at distance r1 to that at r2(= 2 r1) from the wire is ________.
Correct answer is '4 to 4'. Can you explain this answer?
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An infinitely long straight wire is carrying a steady current I. The r...
**Explanation:**
To find the ratio of magnetic energy density at distance r1 to that at distance r2 from the wire, we need to consider the magnetic field produced by the current-carrying wire at these distances.
**1. Magnetic Field at Distance r1:**
The magnetic field produced by an infinitely long straight wire at a distance r from the wire is given by Ampere's Law:
B = (μ₀I)/(2πr)
where B is the magnetic field, μ₀ is the permeability of free space, I is the current, and r is the distance from the wire.
**2. Magnetic Energy Density at Distance r1:**
The magnetic energy density (u) at a point is given by the equation:
u = (B²)/(2μ₀)
Substituting the expression for B, we get:
u₁ = [(μ₀I)/(2πr₁)]² / (2μ₀)
= (μ₀I)² / (4π²r₁² * 2μ₀)
= (μ₀I)² / (8π²r₁²)
**3. Magnetic Field at Distance r2(= 2r1):**
The magnetic field at distance r2 is:
B₂ = (μ₀I)/(2πr₂)
= (μ₀I)/(2π(2r₁))
= (μ₀I)/(4πr₁)
**4. Magnetic Energy Density at Distance r2:**
The magnetic energy density at distance r2 is:
u₂ = [(μ₀I)/(4πr₁)]² / (2μ₀)
= (μ₀I)² / (16π²r₁² * 2μ₀)
= (μ₀I)² / (32π²r₁²)
**5. Ratio of Magnetic Energy Density:**
To find the ratio of magnetic energy density at r1 to that at r2, we divide u₁ by u₂:
(u₁/u₂) = [(μ₀I)² / (8π²r₁²)] / [(μ₀I)² / (32π²r₁²)]
= (μ₀I)² / (8π²r₁²) * (32π²r₁² / (μ₀I)²)
= 4
Therefore, the ratio of magnetic energy density at distance r1 to that at distance r2 is 4 to 4.
To find the ratio of magnetic energy density at distance r1 to that at distance r2 from the wire, we need to consider the magnetic field produced by the current-carrying wire at these distances.
**1. Magnetic Field at Distance r1:**
The magnetic field produced by an infinitely long straight wire at a distance r from the wire is given by Ampere's Law:
B = (μ₀I)/(2πr)
where B is the magnetic field, μ₀ is the permeability of free space, I is the current, and r is the distance from the wire.
**2. Magnetic Energy Density at Distance r1:**
The magnetic energy density (u) at a point is given by the equation:
u = (B²)/(2μ₀)
Substituting the expression for B, we get:
u₁ = [(μ₀I)/(2πr₁)]² / (2μ₀)
= (μ₀I)² / (4π²r₁² * 2μ₀)
= (μ₀I)² / (8π²r₁²)
**3. Magnetic Field at Distance r2(= 2r1):**
The magnetic field at distance r2 is:
B₂ = (μ₀I)/(2πr₂)
= (μ₀I)/(2π(2r₁))
= (μ₀I)/(4πr₁)
**4. Magnetic Energy Density at Distance r2:**
The magnetic energy density at distance r2 is:
u₂ = [(μ₀I)/(4πr₁)]² / (2μ₀)
= (μ₀I)² / (16π²r₁² * 2μ₀)
= (μ₀I)² / (32π²r₁²)
**5. Ratio of Magnetic Energy Density:**
To find the ratio of magnetic energy density at r1 to that at r2, we divide u₁ by u₂:
(u₁/u₂) = [(μ₀I)² / (8π²r₁²)] / [(μ₀I)² / (32π²r₁²)]
= (μ₀I)² / (8π²r₁²) * (32π²r₁² / (μ₀I)²)
= 4
Therefore, the ratio of magnetic energy density at distance r1 to that at distance r2 is 4 to 4.
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An infinitely long straight wire is carrying a steady current I. The ratio of magnetic energy density at distance r1 to that at r2(= 2 r1) from the wire is ________.Correct answer is '4 to 4'. Can you explain this answer?
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