Equating magnetic force to centripetal force,
$$\frac{{m{v^2}}}{r} = qvB\sin {90^ \circ }$$
Time to complete one revolution,
$$T = \frac{{2\pi r}}{v} = \frac{{2\pi m}}{{qB}}$$

$$\eqalign{ & V = {i_g}\left( {G + R} \right) = 4 \times {10^{ - 3}} \cr & \left( {50 + 5000} \right) = 20V \cr} $$

According to Ampere’s circuital law,
$$\int {B \cdot dl = {\mu _0}{i_{{\text{enclosed}}}}} $$
So, $$B\left( {2\pi r} \right) = {\mu _0} \times 0\,\,\left[ {{i_{{\text{enclosed}}}} = 0} \right]$$
$$\therefore B = 0$$
So, inside a hollow metallic (copper) pipe there is no current inside the Ampere’s surface so, the magnetic field is zero.
But for external points, the whole current behaves as if it were concentrated at the axis only, so outside
$${B_0} = \frac{{{\mu _0}i}}{{2\pi r}}$$
Thus, the magnetic field is produced outside the pipe only.

Use the vector form of $$B$$ and $$v$$ in the formulae $$\overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right)$$    to get the instantaneous direction of force at $$x = a$$  and $$x = 2a.$$

As we know, radius of circular path in magnetic field
$$r = \frac{{\sqrt {2Km} }}{{qB}}$$
For electron, $${r_e} = \frac{{\sqrt {2K{m_e}} }}{{eB}}\,......\left( {\text{i}} \right)$$
For proton, $${r_p} = \frac{{\sqrt {2K{m_p}} }}{{eB}}\,......\left( {{\text{ii}}} \right)$$
For $$\alpha $$ particle,
$$\eqalign{ & {r_\alpha } = \frac{{\sqrt {2K{m_a}} }}{{{q_\alpha }B}} = \frac{{\sqrt {2K4{m_p}} }}{{2eB}} = \frac{{\sqrt {2K{m_p}} }}{{eB}}\,......\left( {{\text{iii}}} \right) \cr & \therefore {r_e} < {r_p} = {r_\alpha }\,\,\,\,\,\left( {\because {m_e} < {m_p}} \right) \cr} $$

The change in K.E. is equal to work done by net force which is zero because the magnetic force is perpendicular to velocity. K.E. remains constant.

The time period of electron moving in a circular orbit,
$$T = \frac{{{\text{Circumference of circular path}}}}{{{\text{Speed}}}} = \frac{{2\pi r}}{v}$$
Now, equivalent current due to flow of electron is given by
$$i = \frac{q}{T} = \frac{e}{{\left( {\frac{{2\pi r}}{v}} \right)}} = \frac{{ev}}{{2\pi r}}\,\,\left[ {q = e} \right]$$
Magnetic field at centre of circle
$$\eqalign{ & B = \frac{{{\mu _0}i}}{{2r}} = \frac{{{\mu _0}ev}}{{4\pi {r^2}}}\,\,\,\left( {i = \frac{{eV}}{{2\pi r}}} \right) \cr & \Rightarrow r \propto \sqrt {\frac{V}{B}} \cr} $$

$$\eqalign{ & B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2\pi i}}{r}\,\,{\text{where}} \cr & i = \frac{{2e}}{t} = \frac{{2 \times 1.6 \times {{10}^{ - 19}}}}{2} = 1.6 \times {10^{ - 19}}\,A \cr & \therefore B = \frac{{{\mu _0}i}}{{2r}} = \frac{{{\mu _0} \times 1.6 \times {{10}^{ - 19}}}}{{2 \times 0.8}} \cr & = {\mu _0} \times {10^{ - 19}}\,T \cr} $$

Radius of circular path of the charged particles $$r = \frac{{mV}}{{qB}}$$
Here $$R < r,$$  hence the particle will press the outer wall of the pipe hence the force applied by the pipe on the particle should be towards the centre of the pipe.

Magnetic field at the centre of loop, $${B_1} = \frac{{{\mu _0}I}}{{2R}}$$
Dipole moment of circular loop is $$m = IA$$
$${m_1} = I.A = I.\pi {R^2}\,\,\,\,\left\{ {R = {\text{Radius of the loop}}} \right\}$$
If moment is doubled (keeping current constant) $$R$$ becomes $$\sqrt 2 R$$
$$\eqalign{ & {m_2} = I.\pi {\left( {\sqrt 2 R} \right)^2} = 2.I\pi {R^2} = 2{m_1} \cr & {B_2} = \frac{{{\mu _0}I}}{{2\left( {\sqrt 2 R} \right)}} \cr & \therefore \frac{{{B_1}}}{{{B_2}}} = \frac{{\frac{{{\mu _0}I}}{{2R}}}}{{\frac{{{\mu _0}I}}{{2\left( {\sqrt 2 R} \right)}}}} = \sqrt 2 \cr} $$