Answer:
The y-value is z = 0.759 m
Explanation:
From the question we are told that
The position of the first y-axis is [tex]y_1 = 0.300 \ m[/tex]
The current on the first wire is [tex]I_ 1 = 26.0 \ A[/tex]
The force per unit length on each wire is [tex]\frac{F}{l} = 295 \mu N/m = 295 * 10^{-6} \ N/m[/tex]
Generally the force per unit length on first wire is mathematically represented as
[tex]\frac{F}{l} = \frac{\mu_o * I_1 * I_2 }{2*\pi* y_1}[/tex]
Where [tex]\mu _o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting values
[tex]295 *10^{-6} = \frac{ 4\pi * 10^{-7} * 26.0 * I_2 }{2 *3.142* 0.300}[/tex]
[tex]I_2 = \frac{295 *10^{-6 } * 0.300 * 2* 3.142 }{ 4\pi * 10^{-7} * 26 }[/tex]
[tex]I_2 = 17.0 \ A[/tex]
Now the at the point where the magnetic field is zero the magnetic field of each wire are equal , let that point by z meters from the second wire on the y-axis so
[tex]\frac{\mu_o I_2}{2 * \pi * y_1} = \frac{\mu_o I_1}{2 * \pi * (y_1-z)}[/tex]
[tex]I_2 (y_1 - z) = I_1 * y_1[/tex]
substituting values
[tex]17.0 ( 0.300 - z) = 26 * 0.300[/tex]
z = 0.759 m
