Question

Find the self-inductance of a 1700-turn solenoid 55 cm long and 4.0 cm in diameter. Express your answer with the appropriate units.

Answer 1

The self-inductance of the solenoid is 8.25 mH.

The given parameters;

• number of turns, N = 1700 turn
• length of the solenoid, l = 55 cm = 0.55 m
• diameter of solenoid, d = 4 cm
• radius of the solenoid, r = 2 cm = 0.02 m

The area of the solenoid is calculated as follows;

$A = \pi r^2\\\\A = \pi \times (0.02)^2\\\\A = 0.00125 \ m^2$

The self-inductance of the solenoid is calculated as follows;

$L = \frac{\mu_o N^2 A }{l} \\\\L = \frac{(4\pi \times 10^{-7}) \times 1700^2 \times 0.00125}{0.55} \\\\L = 0.00825 \ H\\\\L = 8.25\ mH$

Thus, the self-inductance of the solenoid is 8.25 mH.

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