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 = 4p × 10-7 T · m/A)
The Correct Answer and Explanation is :
Solution:
The winding density ((n)) of a solenoid refers to the number of turns of wire per unit length. The energy stored in a solenoid can be expressed as:
[
U = \frac{1}{2} L I^2
]
where:
- (U = 6.00 \, \mu\text{J} = 6.00 \times 10^{-6} \, \text{J}) is the energy stored,
- (L) is the inductance,
- (I = 0.400 \, \text{A}) is the current.
The inductance of a solenoid is given by:
[
L = \mu_0 n^2 A l
]
where:
- (\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}) is the permeability of free space,
- (n) is the winding density (turns per unit length),
- (A = \pi r^2) is the cross-sectional area ((r = 5.00 \, \text{cm} = 0.0500 \, \text{m})),
- (l = 0.700 \, \text{m}) is the length of the solenoid.
Step 1: Solve for (L) using the energy equation.
[
L = \frac{2U}{I^2}
]
[
L = \frac{2(6.00 \times 10^{-6})}{(0.400)^2} = 7.50 \times 10^{-5} \, \text{H}
]
Step 2: Rearrange the inductance formula to solve for (n).
[
L = \mu_0 n^2 A l \implies n^2 = \frac{L}{\mu_0 A l}
]
[
n = \sqrt{\frac{L}{\mu_0 A l}}
]
Step 3: Substitute values.
[
A = \pi (0.0500)^2 = 7.854 \times 10^{-3} \, \text{m}^2
]
[
n = \sqrt{\frac{7.50 \times 10^{-5}}{(4\pi \times 10^{-7})(7.854 \times 10^{-3})(0.700)}}
]
[
n = \sqrt{\frac{7.50 \times 10^{-5}}{6.902 \times 10^{-9}}} = \sqrt{1.087 \times 10^4} = 104.3 \, \text{turns/m}
]
Final Answer:
The winding density is approximately 104 turns per meter.
Explanation (300 words):
The winding density of a solenoid directly influences its inductance and magnetic field properties. In this problem, the solenoid stores a known amount of energy when a current passes through it. The relationship between the stored energy, current, and inductance allows us to determine the inductance first. Using the formula (U = \frac{1}{2} L I^2), the inductance is calculated as (7.50 \times 10^{-5} \, \text{H}).
Next, the inductance formula for a solenoid, (L = \mu_0 n^2 A l), connects the physical properties of the solenoid (cross-sectional area, length, and winding density) with its inductance. By rearranging this formula, the winding density ((n)) can be isolated. The solenoid’s cross-sectional area is derived from its radius, and the given permeability of free space ((\mu_0)) and length ((l)) are used.
After substituting the values into the formula, the result is approximately (104 \, \text{turns/m}). This indicates that for every meter of the solenoid’s length, there are 104 wire loops. This density ensures the solenoid’s ability to store the specified energy and produce the required inductance for the given current.
Understanding winding density is essential in designing electromagnetic devices, as it determines the strength and efficiency of the magnetic field generated.