Question

Inside a motor, 30.0 Apasses through a 250-turn circular loop that is 10.0 cmin radius. What is the magnetic field strength created at its center?

Short answer

The magnetic field strength calculated is 0.047 T.

Step-by-step solution

Step 1:Given data

The Current inside a motor is$\mathrm{I}=30.0\mathrm{A}\backslash$

The number of turns in the loop is$\mathrm{N}=250\mathrm{turns}$

The radius of the wire is$\mathrm{R}=10.0\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.10\mathrm{m}$

Determination of magnetic field

The magnetic impact of electric charges in relative motion and magnetized objects is described by a magnetic field, which is a vector field.

The following equation can be used to obtain the magnetic field strength created at the center,

$\mathbf{B}\mathbf{=}\left(\frac{{\mathrm{N\mu }}_{0}I}{2R}\right)$…………………….(1)

Where $\mathbf{N}\mathbf{,}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{,}$and R are the number of turns, the magnetic permeability of vacuum, and the radius of the wire, respectively.

Evaluating the magnetic field

Substitute the given data in equation (1), and we get,

$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{N\mu }}_0\mathrm{I}}{2} \\ =\frac{250\times \left(4\mathrm{\pi }\times {10}^{-7}\mathrm{N}/{\mathrm{m}}^2\right)\times 30\mathrm{A}}{2\times 0.10\mathrm{m}} \\ =0.047\mathrm{T} \end{gathered}$$

Therefore, the magnetic field strength is 0.047 T.