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Q62PE

Expert-verifiedFound in: Page 814

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Inside a motor, 30.0 A ****passes through a 250 ****-turn circular loop that is 10.0 cm ****in radius. What is the magnetic field strength created at its center?**

The magnetic field strength calculated is 0.047 T.

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 $\mathrm{R}=10.0\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.10\mathrm{m}$

**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.**

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

$\mathrm{B}=\frac{{\mathrm{N\mu}}_{0}\mathrm{I}}{2}\phantom{\rule{0ex}{0ex}}=\frac{250\times \left(4\mathrm{\pi}\times {10}^{-7}\mathrm{N}/{\mathrm{m}}^{2}\right)\times 30\mathrm{A}}{2\times 0.10\mathrm{m}}\phantom{\rule{0ex}{0ex}}=0.047\mathrm{T}$

Therefore, the magnetic field strength is 0.047 T.

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