The magnetic field varies inversely with the distance for a long conductor. That is, $$B \propto \frac{1}{d}.$$   According to the magnitude and direction shown graph (A) is the correct one.

The force $$F$$ on the charged particle due to external magnetic field provides the required centripetal force $$\left( { = \frac{{m{v^2}}}{r}} \right)$$   necessary for motion along the circular path of radius $$r.$$
So, $$qvB = \frac{{m{v^2}}}{r}\,\,{\text{or}}\,\,r = \frac{{mv}}{{qB}}$$
$$\therefore r \propto v\,\,\left[ {\frac{m}{{qB}} = {\text{constant}}} \right]$$
As $$v$$ is doubled, the radius also becomes double.
Hence, radius $$ = 2 \times 2 = 4\,cm.$$

Magnetic field due to segment '1'
$$\eqalign{ & \overrightarrow {{B_1}} = \frac{{{\mu _0}I}}{{4\pi R}}\,\left[ {\sin {{90}^ \circ } + \sin {0^ \circ }} \right]\left( { - \hat k} \right) \cr & = \frac{{ - {\mu _0}I}}{{4\pi R}}\left( {\hat k} \right) = \overrightarrow {{B_3}} \cr} $$
Magnetic field due to segment '2'
$$\eqalign{ & {B_2} = \frac{{{\mu _0}I}}{{4\pi R}}\left( { - \hat i} \right) = \frac{{ - {\mu _0}I}}{{4\pi R}}\left( {\pi \hat i} \right) \cr & \therefore \overrightarrow B \,{\text{at}}\,{\text{centre}} \cr & \overrightarrow {{B_c}} = \overrightarrow {{B_1}} + \overrightarrow {{B_2}} + \overrightarrow {{B_3}} \cr & = - \frac{{{\mu _0}I}}{{4\pi R}}\left( {\pi \hat i + 2\hat k} \right) \cr} $$

Here, current is uniformly distributed across the cross-section of the wire, therefore, current enclosed in the amperean path formed at a distance $${r_1}\left( { = \frac{a}{2}} \right) = \left( {\frac{{\pi r_1^2}}{{\pi {a^2}}}} \right) \times I,$$      where $$I$$ is total current
∴ Magnetic field at $${P_1}$$ is
$$\eqalign{ & {B_1} = \frac{{{\mu _0} \times {\text{current enclosed}}}}{{{\text{Path}}}} \cr & \Rightarrow {B_1} = \frac{{{\mu _0} \times \left( {\frac{{\pi r_1^2}}{{\pi {a^2}}}} \right) \times I}}{{2\pi {r_1}}} = \frac{{{\mu _0} \times I{r_1}}}{{2\pi {a^2}}} \cr} $$
Now, magnetic field at point $${P_2},$$
$${B_2} = \frac{{{\mu _0}}}{{2\pi }}.\frac{I}{{\left( {2a} \right)}} = \frac{{{\mu _0}I}}{{4\pi {a^2}}}.$$
∴ Required ratio = $$\frac{{{B_1}}}{{{B_2}}} = \frac{{{\mu _0}I{r_1}}}{{2\pi {a^2}}} \times \frac{{4\pi a}}{{{\mu _0}I}}$$
$$ = \frac{{2{r_1}}}{a} = \frac{{2 \times \frac{a}{2}}}{a} = 1.$$

$$\eqalign{ & r = \frac{{mv}}{{qB}} \cr & \sin \theta = \frac{x}{r} = \frac{{\frac{{mv}}{{\sqrt 2 qB}}}}{{\frac{{mv}}{{qB}}}} = \frac{1}{{\sqrt 2 }} \cr & {\text{or}}\,\,\theta = \frac{\pi }{4} \cr} $$
Time to complete the circle $$\left( {2\pi } \right),T = \frac{{2\pi m}}{{qB}}$$
∴ time taken to traverses $$\frac{\pi }{4},t = \frac{{\pi m}}{{4qB}}$$
Time taken to travel horizontal distance
$${t_1} = \frac{{\frac{{mv}}{{\sqrt 2 qB}}}}{{\frac{v}{{\sqrt 2 }}}} = \frac{m}{{qB}}$$
Total time taken $$ = 2t + 2{t_1} = \frac{m}{{2qB}}\left( {\pi + 4} \right)$$

$$\eqalign{ & {I_g} = \frac{{10 \times {{10}^{ - 3}}}}{5} = 2 \times {10^{ - 3}}\,A; \cr & R = \frac{V}{{{I_g}}} - {R_g} = \frac{1}{{2 \times {{10}^{ - 3}}}} - 5 = 495\,\Omega . \cr} $$

The coil must orient so that its magnetic moment becomes parallel to the field. So that the magnetic force on the coil is zero.

Due to torque of magnetic field, ring will rotate about vertical diameter. $$\tau = I\alpha \Rightarrow MB = I\alpha $$
$$\eqalign{ & \Rightarrow i\pi {r^2}B = \frac{1}{2}m{r^2}\alpha \cr & \Rightarrow \alpha = \frac{{2iB\pi }}{m} = \frac{{2 \times 4 \times 10\pi }}{2} = 40\,\pi \,rad\,{s^{ - 2}} \cr} $$

Applying right hand grip rule and considering $$AB,$$  the direction of magnetic field due to one current is upwards and that due to other is downwards. Both the magnetic fields cancel out each other and the resultant magnetic field is zero.
Considering $$CD$$  and applying right hand grip rule for the two currents, the direction of magnetic field is in the same direction in both the cases giving non-zero resultant.

$$\eqalign{ & F = Bi\ell \cr & 2T\sin \frac{\theta }{2} = F \cr & {\text{or}}\,\,2T\left( {\frac{\theta }{2}} \right) = Bi\left( {r\theta } \right) \cr & \therefore T = Bir \cr & {\text{Now speed, }}v = \sqrt {\frac{T}{m}} \cr & = \sqrt {\frac{{Bir}}{m}} \cr} $$