The figure below shows wire 1 in cross section; the wire is long and straight, c
ID: 2239970 • Letter: T
Question
The figure below shows wire 1 in cross section; the wire is long and straight, carries a current of 3.80 mA out of the page, and is at distance d1 = 2.40 cm from a surface. Wire 2, which is parallel to wire 1 and also long, is at horizontal distance d2 = 5.20 cm from wire 1 and carries a current of 6.80 mA into the page. What is the x component of the magnetic force per unit length on wire 2 due to the current in wire 1?
The figure below shows wire 1 in cross section; the wire is long and straight, carries a current of 3.80 mA out of the page, and is at distance d1 = 2.40 cm from a surface. Wire 2, which is parallel to wire 1 and also long, is at horizontal distance d2 = 5.20 cm from wire 1 and carries a current of 6.80 mA into the page. What is the x component of the magnetic force per unit length on wire 2 due to the current in wire 1?Explanation / Answer
I1= 3.80 mA,d1= 2.40 cm, d2=5.20 cm, I2= 6.80 mA,
Find the x component of the magnetic force per unit lengthon wire 2 due to the current in wire 1.
magnetic field on wire 2 due to wire 1 is
B = ?0I1/(2?d) where d =?(d12+ d22)
the magnetic force per unit length on wire 2 due to the current inwire 1 is
F = BI2
its x component is
Fx= F*d2/d =?0I1I2d2/(2?d2) =?0I1I2d2/[2?(d12+ d22)] = 8.19*10-11N/m