Initial acceleration is given by
$$\eqalign{ & {a_0} = \frac{{eE}}{m}\,\,\left( {F = eE} \right) \cr & \Rightarrow E = \frac{{{a_0}m}}{e}\,\,......\left( {\text{i}} \right) \cr & \therefore \frac{{e{v_0}B + eE}}{m} = 3{a_0} \cr & {\text{or}}\,\,e{v_0}B + eE = 3{a_0}m \cr} $$

$$\eqalign{ & \therefore e{v_0}B = 3{a_0}m - eE \cr & \Rightarrow e{v_0}B = 3m{a_0} - m{a_0}\,\,\left[ {{\text{from Eq}}{\text{.}}\left( {\text{i}} \right)} \right] \cr} $$
$$\therefore B = \frac{{2m{a_0}}}{{e{v_0}}},$$    in vertically downward direction

$$\eqalign{ & M = NiA \cr & \Rightarrow M \propto A \cr & \Rightarrow M \propto {r^2}\,\,\left[ {{\text{As}}\,\ell = 2\pi r \Rightarrow \ell \propto r} \right] \cr & \Rightarrow M \propto {\ell ^2} \cr} $$

The particle will not collide if
$$\eqalign{ & d > 2\left( {{r_1} + {r_2}} \right) \cr & {\text{or}}\,\,d > 2\left( {\frac{{m{v_1}}}{{Bq}} + \frac{{m{v_2}}}{{Bq}}} \right)\,\,{\text{or}}\,\,d > \frac{{2m}}{{Bq}}\left( {{v_1} + {v_2}} \right) \cr} $$

As $$0.2\% $$  of main current passes through the galvanometer hence $$\frac{{998}}{{1000}}I$$  current through the shunt.
$$\eqalign{ & \left( {\frac{{2I}}{{1000}}} \right)G = \left( {\frac{{998I}}{{1000}}} \right)S \cr & \Rightarrow S = \frac{G}{{499}} \cr} $$

Total resistance of Ammeter
$$R = \frac{{SG}}{{S + G}} = \frac{{\left( {\frac{G}{{499}}} \right)G}}{{\left( {\frac{G}{{499}}} \right) + G}} = \frac{G}{{500}}$$

Current in a small element, $$dI = \frac{{d\theta }}{\pi }I$$
Magnetic field due to the element
$$dB = \frac{{{\mu _0}}}{{4\pi }}\frac{{2dl}}{R}$$
The component $$dB\cos \theta ,$$  of the field is cancelled by another opposite component.
Therefore,

$${B_{net}} = \int {dB\sin \theta = \frac{{{\mu _0}I}}{{2{\pi ^2}R}}\int\limits_0^\pi {\sin \theta d\theta = } \frac{{{\mu _0}I}}{{{\pi ^2}R}}} $$

$$\eqalign{ & T = \frac{{2\pi m}}{{qB}},{\left( {\frac{m}{q}} \right)_{He}} > {\left( {\frac{m}{q}} \right)_p} > {\left( {\frac{m}{q}} \right)_e} \cr & \therefore {t_{He}} > {t_p} > {t_e} \cr} $$

We know that $$F = - \frac{{dv}}{{dr}}$$   where $$r$$ = distance of the loop from straight current carrying wire

$$\eqalign{ & {\text{Here }}U = - \overrightarrow m .\overrightarrow B = - {I_2}\pi {a^2} \times \frac{{{\mu _0}}}{{4\pi }}\frac{{{I_1}}}{r} \times 2 \times \cos 0 \cr & = - \frac{{{\mu _0}{I_1}{I_2}{a^2}}}{{2r}} \cr & \therefore F = - \frac{d}{{d\left( r \right)}}\left[ { - \frac{{{\mu _0}{I_1}{I_2}{a^2}}}{{2r}}} \right] = - \frac{{{\mu _0}{I_1}{I_2}{a^2}}}{{{r^2}}} \cr & {\text{Here }}r = d \cr & \therefore F \propto \frac{{{a^2}}}{{{d^2}}}\left( {{\text{attractive}}} \right) \cr} $$

$$R = \frac{{mv}}{{qB}};\vec F = q\left( {\vec v \times \vec B} \right)$$
As $$R$$ decreases, $$v$$ will decrease.
Clockwise motion for upward magnetic field $$ \to + ve$$

The magnetic field due to circular coil 1 and 2 are
$$\eqalign{ & {B_1} = \frac{{{\mu _0}{i_1}}}{{2r}} = \frac{{{\mu _0}{i_1}}}{{2\left( {2\pi \times {{10}^{ - 2}}} \right)}} = \frac{{{\mu _0} \times 3 \times {{10}^2}}}{{4\pi }} \cr & {B_2} = \frac{{{\mu _0}{i_2}}}{{2\left( {2\pi \times {{10}^{ - 2}}} \right)}} = \frac{{{\mu _0} \times 4 \times {{10}^2}}}{{4\pi }} \cr & B = \sqrt {B_1^2 + B_2^2} = \frac{{{\mu _0}}}{{4\pi }}.5 \times {10^2} \cr & \Rightarrow B = {10^{ - 7}} \times 5 \times {10^2} \Rightarrow B = 5 \times {10^{ - 5}}Wb/{m^2} \cr} $$

NOTE : If we take individual length for the purpose of calculating the magnetic field in a 3-Dimensional figure then it will be difficult.
Here a smart choice is divide the loop into two loops. One loop is $$ADEFA$$   in $$y-z$$  plane and the other loop will be $$ABCDA$$   in the $$x-y$$  plane.

We actually do not have any current in the segment $$AD.$$  By choosing the loops we find that in one loop we have to take current from $$A$$ to $$D$$ and in the other one from $$D$$ to $$A.$$ Hence these two cancel out the effect of each other as far as creating magnetic field at the concerned point $$P$$ is considered.
The point $$\left( {a,0,a} \right)$$   is in the $$X-Z$$  plane.
The magnetic field due to current in $$ABCDA$$   will be in + ve $$Z$$-direction.
NOTE : Due to symmetry the $$y$$ -components and $$x$$-components will cancel out each other.
Similarly the magnetic field due to current in $$ADEFA$$   will be in $$x$$ -direction.
∴ The resultant magnetic field will be
$$\overrightarrow B = \frac{1}{{\sqrt 2 }}\left( {\hat i + \hat k} \right)$$