Let the number of turns be = n
Radius = r
Length = l
Length of the wire = circumference of n turns of coil
L = n x 2\(\Pi \) r
r = L/2\(\Pi \) r
Maximum torque = nIBA = nIB \(\Pi \) r²
= nIB\(\Pi \)  (2/2\(\Pi \) n)²
= 1/n
For maximum torque n should be minimum
i.e n = 1

r = mv/qB .........(i)
Also P = mv = \(\sqrt { 2mE } \)..........(ii)
By equ (i) and equ (ii)
As the radius is \(r\frac { \sqrt { 2mE } }{ qB } \) proportional to square root of kinetic energy, so if the kinetic energy is halved the radius become \(\sqrt { 1 } /2\) times of its initial value 

As we know for a charge particle moving in a magnetic fieldm the radius of circular Path r = mv/qB
As both the particles have same velocity therefore
r\(\alpha\)/rd = m\(\alpha\) qd/md q\(\alpha\) 
\(\frac { 4\times 1 }{ 2\times 2 } =\frac { 1 }{ 1 } \)

Magnetic field at O due to current through one infinitely long wire is
\(B=\frac { \mu _{ 0 } }{ 4\pi } \times\frac { 2I }{ r } \hat{k}\)(i.e. along + Z-direction)
So, magnetic field at O due to all the wires is
\(B=\frac { \mu _{ 0 } }{ 4\pi } \times2\times2[\frac { 1 }{ 1 } -\frac { 1 }{ 2 } +\frac { 1 }{ 4 } -\frac { 1 }{ 8 } +\frac { 1 }{ 16 } +...] \hat{ k } \\ =\frac { { \mu }_{ 0 } }{ \pi } [(1+\frac { 1 }{ 4 } +\frac { 1 }{ 16 } +...)\hat{ k } -(\frac { 1 }{ 2 } +\frac { 1 }{ 8 } +...)\hat{ k } ]\\ =\frac { { \mu }_{ 0 } }{ \pi } [(\frac { 1 }{ 1-\frac { 1 }{ 4 } } )\hat{ k } \frac { 1 }{ 2 } (\frac { 1 }{ 1-\frac { 1 }{ 4 } } )\hat{ k } ]\\ \Rightarrow B=\frac { { \mu }_{ 0 } }{ \pi } \times\frac { 2 }{ 3 } \hat{ k } =\frac { { \mu }_{ 0 } }{ 4\pi } \times\frac { { { 8 }\hat{k} } }{ 3 } \)​​​​​​