/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 A single-turn wire loop is \(2.0... [FREE SOLUTION] | 91影视

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Chapter 26: Problem 30

A single-turn wire loop is \(2.0 \mathrm{cm}\) in diameter and carries a 650 -mA current. Find the magnetic field strength (a) at the loop center and (b) on the loop axis, \(20 \mathrm{cm}\) from the center.

Short Answer

Expert verified
The magnetic field strength is \(4.1 \times 10^{-3} \, T\) at the loop center and \(2.0 \times 10^{-6}\, T\) on the loop axis, 20 cm from the center.

Step by step solution

01

Identify given Values and Required Formulas

The current \(I\) is 650 mA or 0.65 A, the radius \(r\) is 1.0 cm or 0.01 m, and the distance \(x\) on the loop axis is 20 cm or 0.2 m. The magnetic constant, \(\mu_{0}\), is \(4\pi \times 10^{-7} \, Tm/A\). The required formulas for this problem are: (a) \(B_{center} = \frac{{\mu_{0}I}}{{2r}}\), and (b) \(B_{axis} = \frac{{\mu_{0}Ir^{2}}}{{2(x^{2}+r^{2})^{3/2}}}\)
02

Calculate the Magnetic Field at the Loop Center

Plug the given values into the formula for part (a) to find the magnetic field at the center of the loop: \(B_{center} = \frac{{4\pi \times 10^{-7}\, Tm/A \times 0.65 \, A}}{{2\times 0.01 \, m}} = 4.1 \times 10^{-3} \, T\)
03

Calculate the Magnetic Field on the Loop Axis

Plug the given values into the formula for part (b) to find the magnetic field on the loop axis, 20 cm from the center: \(B_{axis} = \frac{{4\pi \times 10^{-7} \, Tm/A \times 0.65 \, A \times (0.01 m)^{2}}}{{2\left((0.2 m)^{2}+(0.01 m)^{2}\right)^{3/2}}} = 2.0 \times 10^{-6} \, T\)

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