Question

A $\mathbf{200}\mathbf{-}\mathbf{turn}$solenoid having a length of$\mathbf{25}\mathbf{}\mathbf{cm}$and a diameter of$\mathbf{10}\mathbf{}\mathbf{cm}$carries a current of$\mathbf{0}\mathbf{.}\mathbf{29}\mathbf{}\mathbf{A}$. Calculate the magnitude of the magnetic field inside the solenoid.

Short answer

The magnitude of the magnetic field inside the solenoid is $3\times {10}^{-4}\mathrm{T}$.

Step-by-step solution

Listing the given quantities

• $\mathrm{l}=25\mathrm{cm}=0.25\mathrm{m}$
• $\mathrm{d}=10\mathrm{cm}=0.10\mathrm{m}$
• $\mathrm{N}=200$
• $\mathrm{i}=0.29\mathrm{A}$
  

Understanding the concept of magnetic field and solenoid  

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

Formula:

$\mathrm{B}={\mathrm{\mu }}_0\mathrm{in}={\mathrm{\mu }}_0\mathrm{i}\left(\frac{\mathrm{N}}{\mathrm{l}}\right)$

Calculate the magnitude of the magnetic field

We can calculate the magnetic field.

$\begin{array}{c}\mathrm{B}={\mathrm{\mu }}_0\mathrm{in}\\ ={\mathrm{\mu }}_0\mathrm{i}\left(\frac{\mathrm{N}}{\mathrm{l}}\right)\\ =1.26\times {10}^{-6}\times 0.29\times \frac{200}{0.25}\\ =2.92\times {10}^{-4}\mathrm{T}\\ \approx 3\times {10}^{-4}\mathrm{T}\end{array}$

Hence, the magnitude of the magnetic field inside the solenoid is $3\times {10}^{-4}\mathrm{T}$.