(a) What is the magnitude of the magnetic fieldcreated by I 1 at the location of
ID: 1744706 • Letter: #
Question
(a) What is the magnitude of the magnetic fieldcreated by I1 at the location ofI2?1T 2---Direction---into the monitorin the -x directionout of the monitorin the +x directionin the -y directionin the +y direction
(b) What is the force per unit length exerted onI2 by I1?
3N/m 4---Direction---into the monitorin the -x directionout of the monitorin the +x directionin the -y directionin the +y direction
(c) What is the magnitude of the magnetic field created byI2 at the location ofI1?
5T 6---Direction---in the +x directionin the -y directionout of the monitorin the -x directionin the +y directioninto the monitor
(d) What is the force per length exerted by I2on I1?
7N/m 8---Direction---in the +y directionout of the monitorin the -x directionin the +x directioninto the monitorin the -y direction
Explanation / Answer
given Two long parallel conductors separated by a = 11.0 cm current carried by the first wire I1 = 3.00 A current carried by the second wire I2 =8.00A permeability of free space 0 =4*10-7 Tm/A a) magnitude of the field created by I1 at the location of I2is B1 =oI1/2a = o2I1/4a = (10-7Tm/A)2(3.0A) / 11*10-2m = 0.545*10-5T b) force per unit length exerted by I1 onI2 is F1 = B1I2 = (0.545*10-5T)(8.00 A ) = 43.6T c) = (0.545*10-5T)(8.00 A ) = 43.6T c) magnetic field created by I2 at the location of I1 is B2= oI2/2a = o2I2/4a = (10-7Tm/A)2(8.0A) / 11*10-2m = 1.45*10-5T d) force per unit length exerted by I2 onI1 is F2= B2I1 = (1.45*10-5T)(3A) = 43.5T = (1.45*10-5T)(3A) = 43.5TRelated Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.