Answer:
a
The point on the x-axis where the magnetic field will be zero is [tex] d = 2.8 \ cm [/tex]
b
The point on the x-axis where the magnetic field will be zero is [tex]Z = 5.333 \ cm [/tex]Explanation:
From the question we are told that
The first current is [tex]I_1 = 5.5 \ A[/tex]
The point of x- axis intersection is [tex]x = -2.0 \ cm[/tex]
The second current is [tex]I_2 = 2.5 \ A[/tex]
The point of intersection of the x-axis is [tex]x = 2 .0 \ cm[/tex]
Generally given that the current of the two wires are same direction it means that the magnetic field in -between the wires will cancel out giving zero
So
[tex]B_1 - B_2 = 0[/tex]
=> [tex]B_1 = B_2 [/tex]
=> [tex]\frac{\mu_o * I_1 }{2 * \pi * d } = \frac{\mu_o * I_2 }{2 * \pi * (4 -d) }[/tex]
Here d is the of one wire to the point where the magnetic field is 0
and given that the total distance in-between the wire is D = 2 = 2 = 4
Hence the distance of the other wire to the point where magnetic field is zero is (4 - d)
So
[tex]\frac{5.5}{ d} = \frac{2.5}{4-d}[/tex]
=> [tex] 8d = 22 [/tex]
=> [tex] d = 2.8 \ cm [/tex]
So the point on the x-axis where the magnetic field will be zero is [tex] d = 2.8 [/tex]
Generally given that the current of the two wires are opposition direction it means that the magnetic field at a position which is not in-between the wire will be zero
Let that position be k
Let the distance from the middle of both wires to k be Z
So
[tex]\frac{\mu_o * I_1 }{2 * \pi * ( 2 + Z) } = \frac{\mu_o * I_2 }{2 * \pi * (Z -2) }[/tex]
=> [tex]5.5 Z - 11 = 5+ 2.5 Z[/tex]
=> [tex]3Z = 16[/tex]
=> [tex]Z = 5.333\ cm[/tex]
