Question

A solenoid that is $\mathbf{95}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$long has a radius of$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$and a winding ofrole="math" localid="1663136568607" $\mathbf{1200}\mathbf{}\mathbf{turns}$; it carries a current of$\mathbf{3}\mathbf{.}\mathbf{60}\mathbf{}\mathbf{A}$. Calculate the magnitude of the magnetic field inside the solenoid.

Short answer

The magnitude of the magnetic field inside the solenoid is $0.00571\mathrm{T}.$

Step-by-step solution

Listing the given quantities

$\mathrm{l}=95\mathrm{cm}=0.95\mathrm{m}$

$\mathrm{N}=1200$

$\mathrm{i}=3.60\mathrm{A}$

Understanding the concept of magnetic field and toroid 

The magnetic field produced by the solenoid is given as,

$\mathbf{B}\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{in}\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}\left(\frac{N}{l}\right)$

Here, B is the magnetic field, i is current, n is the number of turns per unit length, N is the total number of turns and l is the length of the solenoid.

We can calculate the magnetic field due to an ideal solenoid. In our case, the solenoid is having a definite length, so we use this equation to calculate the magnetic field inside the solenoid.

Calculating the magnetic field inside the solenoid.

By using the equation,we can calculate the magnetic field.

$\begin{array}{c}\mathrm{B}={\mathrm{\mu }}_0\mathrm{in}\\ ={\mathrm{\mu }}_0\mathrm{i}\left(\frac{\mathrm{N}}{\mathrm{l}}\right)\\ =1.26\times {10}^{-6}\times 3.60\times \frac{1200}{0.95}\\ =0.00571\mathrm{T}\end{array}$

Therefore, the magnitude of the magnetic field inside the solenoid is $0.00571\mathrm{T}$.