For a solenoid magnetic field is given by $$B = {\mu _0}ni$$
where, $$n =$$  number of turns per unit length and
$$i =$$  current through the coil
or so for two different cases $$B \propto ni$$
$$\therefore \frac{{{B_1}}}{{{B_2}}} = \frac{{{n_1}{i_1}}}{{{n_2}{i_2}}}$$
Here, $${n_1} = n,\,{n_2} = \frac{n}{2},$$
$${i_1} = i,{i_2} = 2i,{B_1} = B$$
Hence, $$\frac{B}{{{B_2}}} = \frac{n}{{\frac{n}{2}}} \times \frac{i}{{2i}} = 1\,\,{\text{or}}\,\,{B_2} = B$$

Let us consider an elemental length dl subtending an angle $$d\theta $$  at the centre of the circle. Let $${F_B}$$  be the magnetic force acting on this length. Then

$${F_B} = BI\left( {dl} \right)$$   directed upwards as shown
$$\eqalign{ & = BI\left( {Rd\theta } \right)\,\,\,\,\,\left[ {\because {\text{angle}}\left( {d\theta } \right) = \frac{{arc\left( {dI} \right)}}{{{\text{radius}}R}}} \right] \cr & = BI\left( {\frac{L}{{2\pi }}} \right)d\theta \,\,\,\,\left[ {\because 2\pi R = L \Rightarrow R = \frac{L}{{2\pi }}} \right] \cr} $$
Let $$T$$ be the tension in the wire acting along both ends of the elemental length as shown. On resolving $$T,$$  we find that the components. $$T\cos \left( {\frac{{d\theta }}{2}} \right)$$   cancel out and the components. $$T\sin \left( {\frac{{d\theta }}{2}} \right)$$   add up to balance $${F_B}.$$
At equilibrium $$2T\sin \left( {\frac{{d\theta }}{2}} \right) = BI\frac{L}{{2\pi }}d\theta $$
$$\eqalign{ & \Rightarrow 2T\frac{{d\theta }}{2} = BI\frac{L}{{2\pi }}d\theta \quad \left[ {\because \frac{{d\theta }}{2} = {\text{small}}} \right] \cr & \Rightarrow T = \frac{{BIL}}{{2\pi }} \cr} $$

Magnetic dipole moment
$$\eqalign{ & m = iA = \frac{e}{T} \times \pi {r^2} = \frac{e}{{\left( {\frac{{2\pi r}}{v}} \right)}} \times \pi {r^2} = \frac{{erv}}{2}. \cr & = \frac{{1.6 \times {{10}^{ - 19}} \times 50 \times {{10}^{ - 12}} \times 2.2 \times {{10}^6}}}{2} \cr & = 8.8 \times {10^{ - 25}}A{m^2}. \cr} $$

Magnetic field due to current in wire 1 at point $$P$$ distant $$r$$ from the wire is
$$\eqalign{ & B = \frac{{{\mu _0}}}{{4\pi }}\frac{{{i_1}}}{r}\left[ {\cos \theta + \cos \theta } \right] \cr & B = \frac{{{\mu _0}}}{{2\pi }}\frac{{{i_1}\cos \theta }}{r}\,\left( {{\text{directed perpendicular to the plane of paper, inwards}}} \right) \cr} $$
The force exerted due to this magnetic field on current element $${i_2}dl$$  is $$dF = {i_2}dl\sin {90^ \circ }$$
$$\therefore dF = {i_2}dl\left[ {\frac{{{\mu _0}}}{{2\pi }}\frac{{{i_1}\cos \theta }}{r}} \right] = \frac{{{\mu _0}}}{{2\pi }}{i_1}{i_2}dl\cos \theta $$

Magnetic field due to straight wire is given by
$$B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2i}}{r}$$
From above expression magnetic field due to current carrying conductor doesn’t depends on the diameter of wire, it only depends on distance of wire from the point and current in the wire so, magnetic field reamaining same for both wires.

Here, $${{\vec F}_{AB}} + {{\vec F}_{BCDA}} = \vec 0$$
$$ \Rightarrow {{\vec F}_{BCDA}} = - {{\vec F}_{AB}} = - \vec F\,\,\left( {\because {F_{AB}} = \vec F} \right)$$

The centre will be at $$C$$ as shown :
Coordinates of the centre are
$$\left( {r\cos {{60}^ \circ }, - r\sin {{60}^ \circ }} \right)$$
where $$r$$ = radius of circle
$$ = \frac{{mv}}{{Bq}} = \frac{{1 \times 1}}{{1 \times 1}} = 1$$
i.e., $$\left( {\frac{1}{2}, - \frac{{\sqrt 3 }}{2}} \right)$$

The magnetic field at a point on the axis of a circular loop at a distance $$x$$ from centre is,
$$\eqalign{ & B = \frac{{{\mu _0}i{a^2}}}{{2{{\left( {{x^2} + {a^2}} \right)}^{\frac{3}{2}}}}}\,\,\,\,B' = \frac{{{\mu _0}i}}{{2a}} \cr & \therefore B' = \frac{{B.{{\left( {{x^2} + {a^2}} \right)}^{\frac{3}{2}}}}}{{{a^3}}} \cr & {\text{Put}}\,x = 4\,\& \,a = 3 \Rightarrow B' = \frac{{54\left( {{5^3}} \right)}}{{3 \times 3 \times 3}} = 250\mu T \cr} $$

$$F = iB\left| {\sin \theta .} \right.$$   This is maximum when $$\sin \theta = 1\,\,{\text{or}}\,\,\theta = \frac{\pi }{2}.$$

The velocity at $$P$$ is in the $$X$$-direction (given).
Let $$\overrightarrow v = k\hat i.$$
After $$P,$$ the positively charged particle gets deflected in the $$x - y$$  plane toward $$- y$$  direction and the path is non-circular.
$$\eqalign{ & {\text{Now, }}\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right) \cr & \Rightarrow \overrightarrow F = q\left[ {k\hat i \times \left( {c\hat k + a\hat i} \right)} \right]\,{\text{for option}}\,\left( {\text{B}} \right) \cr & = q\left[ {kc\hat i \times \hat k + \hat ka\hat i \times \hat i} \right] = kcq\left( { - \hat j} \right) \cr} $$
Since in option (B), electric field is also present $$\overrightarrow E = a\hat i,$$   therefore it will also exert a force in the $$+X$$ direction. The net result of the two forces will be a non-circular path.
Only option (B) fits for the above logic. For other option, we get some other results.