Question

To get an ideal of the order of magnitude of inductance, calculate the self-inductance in henries for a solenoid with 1000 loops of wire wound on a rod 10 cm long with radius 1 cm. If the solenoid were filled with iron so that the actual magnetic field were 10 times larger for the same current in the solenoid, what would be the inductance?

Short answer

The required inductance is $4\times {10}^{-2}\text{V}$.

Step-by-step solution

A concept:

Self-inductance is the tendency of a coil to resist changes in current by itself. Whenever current changes through the coil, they induce an emf that is proportional to the rate of change of current through the coil.

The self-inductance of a solenoid is defined as the flux associated with the solenoid when a unit amount of current passes through it.

The required formula:

The self-induced emf is proportional to the rate of change of the current with proportionality constant$L$ .

$em{f}_{\text{ind}}=L\left|\frac{dI}{dt}\right|$

The proportionality constant $L$is the self-inductance of the solenoid.

Emf induced in the solenoid is,

$emf=\frac{{\mu }_0{N}^2}{d}\pi R^2\frac{dI}{dt}$ ….. (1)

Comparing the above two equations, the self-inductance of the solenoid is,

$L=\frac{{\mu }_0{N}^2}{d}\pi R^2$ ….. (2)

Given data:

Number of loops of wire, $N=1000$

Length of a rod, $d=10\text{cm}=10\times {10}^{-2}\text{m}$

The radius of a rod, $r=1\text{cm}=\text{1}\times {\text{10}}^{-2}\text{m}$

The permeability${\mu }_0=4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$

Define the inductance:

The self-inductance of the solenoid is defined by substituting known values into equation (2).

$\begin{array}{c}L=\frac{{\mu }_0{N}^2}{d}\pi R^2\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right){\left(1000\right)}^2}{10\times {10}^{-2}\text{m}}\times 3.14\times {\left(1\times {10}^{-2}\right)}^2\\ =4\times {10}^{-3}\text{V}\end{array}$

It is given that the magnetic field is increased by 10 times by inserting the iron core. The self-inductance depends on dimensions of the solenoid and material of the core. Since magnetic field is increased by 10 times the permeability$\left(\mu \right)$ of the core is 10 times larger than the previous core. Self-inductance is directly proportional to permeability. So, the self-inductance is increased by the factor of 10.

$\begin{array}{c}L\text{'}=10L\\ =10\left(4\times {10}^{-3}\text{V}\right)\\ =4\times {10}^{-2}\text{V}\end{array}$

Hence, the required inductance is$4\times {10}^{-2}\text{V}$