Chapter 22: Q62PE (page 814)

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?

Expert verified

The magnetic field strength calculated is 0.047 T.

Step by step solution

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

The number of turns in the loop is $N = 250 turns$

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

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,

$B = (\frac{{Nμ}_{0} I}{2 R})$ …………………….(1)

Where $N, μ_{0},$ 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,

$B = \frac{{Nμ}_{0} I}{2} = \frac{250 \times (4 π \times 10^{- 7} N / m^{2}) \times 30 A}{2 \times 0.10 m} = 0.047 T$

Therefore, the magnetic field strength is 0.047 T.

Unlock Step-by-Step Solutions & Ace Your Exams!

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.