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### College Physics (Urone)

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.

See the step by step solution

## Step 1:Given data

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

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

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

## Step 2: 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.

## Step 3: Evaluating the magnetic field

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

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

Therefore, the magnetic field strength is 0.047 T.