Answer:
The energy stored in the solenoid is 7.078 x 10⁻⁵ J
Explanation:
Given;
diameter of the solenoid, d = 2.80 cm
radius of the solenoid, r = d/2 = 1.4 cm
length of the solenoid, L = 14 cm = 0.14 m
number of turns, N = 200 turns
current in the solenoid, I = 0.8 A
The cross sectional area of the solenoid is given as;
[tex]A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2[/tex]
The inductance of the solenoid is given by;
[tex]L = \frac{\mu_0 N^2A}{l} \\\\L = \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H[/tex]
The energy stored in the solenoid is given by;
E = ¹/₂LI²
E = ¹/₂(2.212 x 10⁻⁴)(0.8)²
E = 7.078 x 10⁻⁵ J
Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J
The energy stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800 A is 7.07 × 10⁻⁵ J.
A solenoid is a type of electromagnet, whose purpose is to generate a controlled magnetic field through a coil wound into a tightly packed helix.
First, given the diameter of the solenoid (d) is 2.80 cm (0.0280 m), we can calculate the cross-sectional area (A) using the following expression.
[tex]A =\pi \times (\frac{d}{2} )^{2} = \pi \times (\frac{0.0280}{2} )^{2} = 6.16 \times 10^{-4} m^{2}[/tex]
Next, we will calculate the inductance (L) of the solenoid using the following expression.
[tex]L = \frac{\mu _0 \times N^{2} \times A}{l} = \frac{(4\pi \times 10^{-7}H/m ) \times 200^{2} \times (6.16 \times 10^{-4}m^{2} )}{0.140m} = 2.21 \times 10^{-4} H[/tex]
where,
Finally, for a current (I) of 0.800 A, we can calculate the energy stored (E) using the following expression.
[tex]E = \frac{1}{2} \times L \times I^{2} = \frac{1}{2} \times (2.21 \times 10^{-4}H ) \times (0.800A)^{2} = 7.07 \times 10^{-5} J[/tex]
The energy stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800 A is 7.07 × 10⁻⁵ J.
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