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dlinn [17]
3 years ago
10

To get an idea of the order of magnitude of inductance, calculate the self-inductance in henries for a solenoid with 1500 loops

of wire wound on a rod 13 cm long with radius 2 cm
Physics
1 answer:
valkas [14]3 years ago
8 0

Answer:

The self-inductance in henries for the solenoid is 0.0274 H.

Explanation:

Given;

number of turns, N = 1500 turns

length of the solenoid, L = 13 cm = 0.13 m

radius of the wire, r = 2 cm = 0.02 m

The self-inductance in henries for a solenoid is given by;

L = \frac{\mu_oN^2A}{l}

where;

\mu_o is permeability of free space = 4\pi*10^{-7} \ H/m

A is the area of the solenoid = πr² = π(0.02)² = 0.00126 m²

L = \frac{4\pi *10^{-7}(1500)^2*(0.00126)}{0.13} \\\\L = 0.0274 \ H

Therefore, the self-inductance in henries for the solenoid is 0.0274 H.

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