A solenoid is 1.41 m long and has 485 turns per meter. What is the cross-sectional area of this solenoid if it stores 0.320 J of energy when it carries a current of 14.0 A?
To find the cross-sectional area of the solenoid, we first need to find the magnetic field inside the solenoid. The energy stored in a solenoid with length L, number of turns per unit length n, and current I is given by:
\(U = \frac{1}{2} \mu_0 \frac{N^2 A I^2}{L}\)
where:
- \(U\) is the energy stored in the solenoid (0.320 J),
- \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{Tm/A}\)),
- \(N\) is the total number of turns in the solenoid (\(N = n \times L = 485 \, \text{turns/m} \times 1.41 \, \text{m}\)),
- \(A\) is the cross-sectional area of the solenoid, and
- \(I\) is the current flowing through the solenoid (14.0 A).
First, we find the total number of turns:
\(N = 485 \, \text{turns/m} \times 1.41 \, \text{m} = 683.85 \, \text{turns}\)
Substitute the values into the equation:
\(0.320 = \frac{1}{2} \times 4\pi \times 10^{-7} \times \frac{(683.85)^2 \times A \times (14.0)^2}{1.41}\)
Solving for \(A\):
\(A = \frac{0.320 \, \text{J} \times 1.41 \, \text{m}}{\frac{1}{2} \times 4\pi \times 10^{-7} \times (683.85)^2 \times (14.0)^2}\)
\(A \approx 5.44 \times 10^{-6} \, \text{m}^2\)
Therefore, the cross-sectional area of the solenoid is approximately \(5.44 \times 10^{-6}\, \text{m}^2\).