Answer:
0.67cm
Explanation:
We are given that
Current ,[tex]I_1=5A[/tex]
[tex]I_2=2.5 A[/tex]
[tex]x_1=-2 cm[/tex]
[tex]x_2=2 cm[/tex]
[tex]d=x_2-x_1=2+2=4 cm[/tex]
Let x is the distance between two wires
Magnetic field in the wire,[tex]B=\frac{2\mu_0I}{4\pi r}[/tex]
Where [tex]\frac{\mu_0}{4\pi}=10^{-7}[/tex]
[tex]r_1=x[/tex]
[tex]r_2=4-x[/tex]
[tex]B_1=B_2[/tex]
[tex]\frac{2\mu_0 I_1}{4\pi r_1}=\frac{2\mu_0I_2}{4\pi r_2}[/tex]
[tex]\frac{I_1}{r_1}=\frac{I_2}{r_2}[/tex]
[tex]\frac{5}{x}=\frac{2.5}{4-x}[/tex]
[tex]20-5x=2.5x[/tex]
[tex]20=2.5x+5x=7.5x[/tex]
[tex]x=\frac{20}{7.5}=2.67 cm[/tex]
Distance of 5A wire from the point=-2+2.67=0.67 cm.
Hence, at x=0.67 cm the magnetic field is zero.
