Question

he drawing shows two perpendicular, long, straight wires, both of which lie in the plane of the paper. The current in each of the wires is I = 4.7 A. In the drawing dH = 0.19 m and dV = 0.41 m. Find the magnitudes of the net magnetic fields at points A and B.

Answer 1

Answer:

The magnitudes of the net magnetic fields at points A and B is 2.66 x $10^{-6}$ 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$_{0}$I/2πr

where

B = magnetic field (T)

μ$_{0}$ = 1.26 x $10^{-6}$ (N/$A^{2}$)

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$_{0}$I/2πr

     = (1.26 x $10^{-6}$)(4.7)/(2π)(0.19)

     = 4.96 x $10^{-6}$ T

BV = μ$_{0}$I/2πr

     = (1.26 x  $10^{-6}$)(4.7)/(2π)(0.41)

     = 2.3 x $10^{-6}$ T

hence,

the net magnetic field = BH - BV

                                     = 4.96 x $10^{-6}$ - 2.3 x $10^{-6}$

                                     = 2.66 x $10^{-6}$ T