A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 μJ of energy when a 0.400 A current runs through it. What is the winding density of the solenoid? (μ₀ = 4Ï€10â»â· Tm/A) A) 865 turns/m B) 327 turns/m C) 1080 turns/m D) 104 turns/m E) 472 turns/m
A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 μJ of energy when a 0.400 A current runs through it. What is the winding density of the solenoid? (μ₀ = 4Ï€10â»â· Tm/A) A) 865 turns/m B) 327 turns/m C) 1080 turns/m D) 104 turns/m E) 472 turns/m
The Correct Answer and Explanation is:
To find the winding density (number of turns per unit length, n) of the solenoid, we use the formula for the energy stored in the magnetic field of an ideal solenoid: U=12μ0n2AlI2U = \frac{1}{2} \mu_0 n^2 A l I^2
Where:
- U=6.00 μJ=6.00×10−6 JU = 6.00 \ \mu\text{J} = 6.00 \times 10^{-6} \ \text{J} (energy stored)
- μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \ \text{T·m/A} (permeability of free space)
- nn = winding density (turns per meter) [what we want to find]
- A=πr2A = \pi r^2 = cross-sectional area
- l=0.700 ml = 0.700 \ \text{m} = length of solenoid
- r=5.00 cm=0.0500 mr = 5.00 \ \text{cm} = 0.0500 \ \text{m}
- I=0.400 AI = 0.400 \ \text{A} = current
Step-by-step Calculation:
1. Calculate cross-sectional area AA: A=πr2=π(0.0500)2=π×2.5×10−3=7.854×10−3 m2A = \pi r^2 = \pi (0.0500)^2 = \pi \times 2.5 \times 10^{-3} = 7.854 \times 10^{-3} \ \text{m}^2
2. Rearranging the energy formula to solve for nn: U=12μ0n2AlI2⇒n2=2Uμ0AlI2U = \frac{1}{2} \mu_0 n^2 A l I^2 \Rightarrow n^2 = \frac{2U}{\mu_0 A l I^2}
3. Plug in values: n2=2×6.00×10−6(4π×10−7)×(7.854×10−3)×0.700×(0.400)2n^2 = \frac{2 \times 6.00 \times 10^{-6}}{(4\pi \times 10^{-7}) \times (7.854 \times 10^{-3}) \times 0.700 \times (0.400)^2}
First, calculate the denominator: 4π×10−7×7.854×10−3×0.700×0.16≈1.105×10−94\pi \times 10^{-7} \times 7.854 \times 10^{-3} \times 0.700 \times 0.16 \approx 1.105 \times 10^{-9} n2=1.2×10−51.105×10−9≈10859.73n^2 = \frac{1.2 \times 10^{-5}}{1.105 \times 10^{-9}} \approx 10859.73 n≈10859.73≈104.2n \approx \sqrt{10859.73} \approx 104.2
Final Answer:
104 turns/m(Choice D)\boxed{104 \ \text{turns/m}} \quad \text{(Choice D)}
Explanation
A solenoid stores magnetic energy when current flows through it, and this stored energy depends on the winding density — the number of turns per meter of the coil. The magnetic field inside an ideal solenoid is uniform and given by B=μ0nIB = \mu_0 n I, and the energy stored is proportional to the square of this field.
The energy formula for a solenoid is: U=12μ0n2AlI2U = \frac{1}{2} \mu_0 n^2 A l I^2
This equation reflects how energy increases with greater winding density nn, current II, cross-sectional area AA, and solenoid length ll. Given the energy stored (6.00 µJ), current (0.400 A), and the solenoid’s dimensions, we isolated nn and substituted all known values. Calculating the cross-sectional area from the radius (5.00 cm = 0.0500 m), and converting microjoules to joules, we carefully handled the powers of 10 to maintain precision.
After solving for nn, we found the winding density to be approximately 104.2 turns/m. This matches option D (104 turns/m). Small variations due to rounding are common in such physics problems, but the answer remains accurate within significant figures.
This result tells us that the solenoid has about 104 loops of wire per meter, allowing it to generate the required magnetic field to store the given energy with the specified current.
