magnetism

1. Distance of X from each wire = 25/(2.sin(45) cm = 17.68 cm = 0.1768 m
Magnetic flux density at X from each wire = (uo).(2.5)/(2.pi x 0.1768) T = 2.828x10^-6 T
where (uo) is the permeability of free space (= 4.pi x 10^-7 H/m)
Hence, resultant magnetic flux density = square-root[ (2x2.828x10^-6)^2 + (2x2.828x10^-6)^2] T = 8 x10^-6 T

By Right-Hand grip Rule, the resultant magnetic flux density is pointing upward.

2. (a) zero
(b) Magnetic flux density at Q
= [(uo).(1.25)/(2.pi x 0.75) + (uo).(1.25)/(2.pi x 0.25)] T
= 1.333x10^-6 T
[where uo is the permeability of free space]
The direction of the flux density is pointing into the paper.

3. Consider a point with coordinates (x,y) in the first quadrant where the resultant magnetic flux density is zero.
Magnetic flux density at the point due to the 4A current
= (uo).(4)/(2.pi.x)
Magnetic flux density at the point due to the 2A current
= (uo)(2)/(2.pi.y)
Hence, for flux density equals to zero,
(uo)(4)/(2.pi.x) = (uo)(2)/2.pi.y)
i.e. 2/x = 1/y
or 2y = x
y = x/2

(a) The straight line of neutral flux density is represented by y = x/2

(b) The slope of the line = 1/2
hence, angle between the line and x-axis = arc-tan(1/2) degrees = 26.57 degrees