A solenoid of length 0.700m having a circular cross-section of radius 5.00cm stores 6.00 µJ of energy when a 0.400-A current runs through it. What is the winding density of the solenoid? (µ0 = 4p10-7 Tm/A)
A) 865 turns/m
B) 327 turns/m
C) 1080 turns/m
D) 104 turns/m
E) 472 turns/m&nb…
The correct answer and explanation is:
To solve this problem, we will determine the winding density of the solenoid based on the given information.
Key Equations and Variables:
- Energy Stored in a Solenoid: U=12LI2U = \frac{1}{2} L I^2 where UU is the energy stored, LL is the inductance, and II is the current.
- Inductance of a Solenoid: L=μ0n2AℓL = \mu_0 n^2 A \ell where nn is the winding density (turns per unit length), A=πr2A = \pi r^2 is the cross-sectional area of the solenoid, ℓ\ell is its length, and μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A} is the permeability of free space.
Known Values:
- Length of the solenoid: ℓ=0.700 m\ell = 0.700 \, \text{m}
- Radius: r=5.00 cm=0.050 mr = 5.00 \, \text{cm} = 0.050 \, \text{m}
- Energy stored: U=6.00 μJ=6.00×10−6 JU = 6.00 \, \mu\text{J} = 6.00 \times 10^{-6} \, \text{J}
- Current: I=0.400 AI = 0.400 \, \text{A}
Step 1: Calculate LL using the energy equation.
Rearranging the energy equation: L=2UI2L = \frac{2U}{I^2}
Substitute the values: L=2×6.00×10−6(0.400)2=1.20×10−50.16=7.50×10−5 HL = \frac{2 \times 6.00 \times 10^{-6}}{(0.400)^2} = \frac{1.20 \times 10^{-5}}{0.16} = 7.50 \times 10^{-5} \, \text{H}
Step 2: Solve for nn using the inductance equation.
Rearranging the inductance equation: n=Lμ0Aℓn = \sqrt{\frac{L}{\mu_0 A \ell}}
Substitute A=πr2=π(0.050)2=7.85×10−3 m2A = \pi r^2 = \pi (0.050)^2 = 7.85 \times 10^{-3} \, \text{m}^2, μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7}, ℓ=0.700 m\ell = 0.700 \, \text{m}, and L=7.50×10−5L = 7.50 \times 10^{-5}: n=7.50×10−5(4π×10−7)(7.85×10−3)(0.700)n = \sqrt{\frac{7.50 \times 10^{-5}}{(4\pi \times 10^{-7}) (7.85 \times 10^{-3}) (0.700)}}
First, calculate the denominator: 4π×10−7×7.85×10−3×0.700=6.90×10−94\pi \times 10^{-7} \times 7.85 \times 10^{-3} \times 0.700 = 6.90 \times 10^{-9}
Now calculate nn: n=7.50×10−56.90×10−9=1.09×104≈1044 turns/mn = \sqrt{\frac{7.50 \times 10^{-5}}{6.90 \times 10^{-9}}} = \sqrt{1.09 \times 10^{4}} \approx 1044 \, \text{turns/m}
Step 3: Match to the closest answer.
The closest option is C) 1080 turns/m.
Explanation:
The winding density of a solenoid represents the number of turns of wire per unit length. Using the energy stored in the solenoid, we calculated the inductance, which depends on the winding density, cross-sectional area, and length. By rearranging and solving, we found the winding density to be approximately 1044 turns/m1044 \, \text{turns/m}, which aligns closely with option CC. Accurate understanding of energy storage and electromagnetic principles is key to solving this.