Question

What is the self-inductance of a solenoid 30.0 cm long having 100 turns of wire and a cross-sectional area of 1.00 × 10-4 m2? (μ0 = 4π × 10-7 T • m/A)

Answer 1

Answer:

$L=4.19*10^{-6}H$

Explanation:

The self-inductance of a solenoid is defined as:

$L=\frac{\mu N^2A}{l}$

Here $\mu_0$ is the the permeability of free space, N is the number of turns in the solenoid, A is the cross-sectional area os the solenoid and l its length. We replace the given values to get the self-inductance:

$L=\frac{4\pi*10^{-7}\frac{T\cdot m}{A}(100)^2(1*10^{-4}m^2)}{30*10^{-2}m}\\L=4.19*10^{-6}H$

Answer 2

The self-inductance of a solenoid is $4.19 \times 10^{-6}H$

The formula for calculating the self-inductance is expressed as:

$L = \frac{\mu N^2A}{l}$

where:

$\mu_0$ is the permeability of free space

N is the number of turns in the solenoid

A is the cross-sectional area of the solenoid

l is its length.

Given the following parameters:

$\mu_0 = 4\pi \times 10^{-7} T \cdot m/A$

N = 100 turns

A = $1.0 \times 10^{-4}m^2$

l = 30.0cm = 0.3m

Substitute the given parameters into the formula

$L=\frac{4\pi \times 10^{-7} \times 100^2 \times 1.00\times 10^{-4}}{0.3}\\L =4.19 \times 10^{-6}H$

Hence the self-inductance of a solenoid is $4.19 \times 10^{-6}H$

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