Answer:
The magnitudes of the net magnetic fields at points A and B is 2.66 x [tex]10^{-6}[/tex] T
Explanation:
Given information :
The current of each wires, I = 4.7 A
dH = 0.19 m
dV = 0.41 m
The magnetic of straight-current wire :
B= μ[tex]_{0}[/tex]I/2πr
where
B = magnetic field (T)
μ[tex]_{0}[/tex] = 1.26 x [tex]10^{-6}[/tex] (N/[tex]A^{2}[/tex])
I = Current (A)
r = radius (m)
the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,
BH = μ[tex]_{0}[/tex]I/2πr
= (1.26 x [tex]10^{-6}[/tex])(4.7)/(2π)(0.19)
= 4.96 x [tex]10^{-6}[/tex] T
BV = μ[tex]_{0}[/tex]I/2πr
= (1.26 x [tex]10^{-6}[/tex])(4.7)/(2π)(0.41)
= 2.3 x [tex]10^{-6}[/tex] T
hence,
the net magnetic field = BH - BV
= 4.96 x [tex]10^{-6}[/tex] - 2.3 x [tex]10^{-6}[/tex]
= 2.66 x [tex]10^{-6}[/tex] T
