Chapter 20, Problem 4P

To determine

Magnetic field of straight wire carrying current 6.45 A and maximum force acting is 1.28N.

Answer to Problem 4P

Solution:

B=0.357 T

Explanation of Solution

Given:

Current I = 6.45 A; Maximum Force F = 1.28N and l = 55.5 cm

Formula Used:

$F=IlB\mathrm{Sin}\theta$

Calculation:

In order to find the value of Magnetic field of current carrying conductor, we have to use the equation,

$F=IlB\mathrm{Sin}\theta$

Since force in maximum when Ϙ = 90⁰

$F_{\max }=ilB$

$B=\frac{F_{\max }}{il}$

$B=\frac{1.28}{(6.45)(.555)}$

$B=0.357T$