A long horizontal wire carries a current of I 1 = 11.5 A. A second wire, made of
ID: 1265166 • Letter: A
Question
A long horizontal wire carries a current of I1 = 11.5 A. A second wire, made of 0.100 mm diameter copper wire, is 63.0 cm below the first wire and is held in suspension magnetically. What is the magnitude and direction of the current in the bottom wire? Give a + sign if I2 is parallel to I1, and a - sign if I2 is opposite to I1.
The mass density of Copper is 8.93
Explanation / Answer
Magnetic field due to the current carrying conductor, B = u*i1/(2pi*r) = 4pi*10^-7* 11.5 / (2pi*0.63) = 0.00000365079 Telsa
F = i2*length of second wire*B = mass of wire* g
or, i2 = (density*pi*0.0001^2)*g/B = 768.45 A
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.