There are excess surface charges on each wire (irrespective of whether the current is flowing through them or not). Hence in addition to the magnetic force
F

m
​
, we must take into account the electric force
F

e
​
. Suppose that an excess charge λ corresponds to a unit length of the wire, then electric force exerted per unit length of the wire by other wire can be found with the help of Gauss' theorem.
F
e
​
=λE=λ
4πε
0
​

1
​

l
2λ
​
=
4πε
0
​
l
2λ
2

​
, (1)
where l is the distance between the axes of the wires. The magnetic force acting per unit length of the wire can be found with the help of the theorem on circulation of vector
B

F
m
​
=
4π
μ
0
​

​

l
2i
2

​
,
where i is the current in the wire (2)
Now, from the relation,
λ=Cφ, where C is the capacitance of the wires per unit lengths and φ=iR
λ=
lnη
πε
0
​

​
iR or,
λ
i
​
=
πε
0
​
R
lnη
​
(3)
Dividing (2) by (1) and then substituting the value of
λ
i
​
from (3), we get,
F
e
​

F
m
​

​
=
ε
0
​

μ
0
​

​

π
2
R
2

(lnη)
2

​

The resultant force of interaction vanishes when this ratio equals unity. This is possible when R=R
0
​
, where
R
0
​
=
ε
0
​

μ
0
​

​

​

π
lnη
​
=0.36kΩ