The force acting on electron will be perpendicular to the direction of velocity till the electron remains in the magnetic field. So the electron will follow the path as given.

A galvanometer can be converted into an ammeter by using a low resistance wire in parallel with the galvanometer. The resistance of this wire depends upon the range of the ammeter and it is given by

Shunt resistance $$S = \left( {\frac{{{i_g}G}}{{i - {i_g}}}} \right)$$

Apply Ampere's circular law to the coaxial circular loops $${L_1}$$ and $${L_2}.$$ The magnetic field is $${B_1}$$ at all points on $${L_1}$$ and $${B_2}$$ at all points on $${L_2}.$$ $$\sum I \ne 0$$   for $${L_1}$$ and 0 for $${L_2}.$$
Hence, $${B_1} \ne 0$$  but $${B_2} = 0$$
$$\left[ {{\text{As}}\,\oint {\vec B.d\vec i = {\mu _0}\sum I} } \right]$$

The angular momentum $$L$$ of the particle is given by $$L = m{r^2}\omega \,{\text{where}}\,\,\omega = 2\pi n.$$
$$\eqalign{ & \therefore {\text{Frequency}}\,n = \frac{\omega }{{2\pi }};{\text{Further}}\,i = q \times n = \frac{{\omega q}}{{2\pi }} \cr & {\text{Magnetic}}\,{\text{moment,}}\,M = iA = \frac{{\omega q}}{{2\pi }} \times \pi {r^2}; \cr & \therefore M = \frac{{\omega q{r^2}}}{2}\,{\text{so,}}\,\frac{M}{L} = \frac{{\omega q{r^2}}}{{2m{r^2}\omega }} = \frac{q}{{2m}} \cr} $$

Magnetic field due to a long current carrying wire at distance $$r$$ at point $$P$$ is given by
$$B = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{2i}}{r}$$
So, $$B \propto \frac{1}{r}$$
when $$r$$ is doubled, the magnetic field becomes halved.
i.e., $$B' = \frac{B}{2} \Rightarrow B' = \frac{{0.4}}{2} = 0.2\,T$$

Magnetic force = centripetal force
i.e. $$qvB = \frac{{m{v^2}}}{r}$$
or $$qvB = mr{\omega ^2}\,\,\left( {v = r\omega } \right)$$
or $${\omega ^2} = \frac{{qvB}}{{mr}} = \frac{{q\left( {r\omega } \right)B}}{{mr}}$$
Angular frequency, $$\omega = \frac{{qB}}{m}$$
If $$\nu $$ is the frequency of rotation, then
$$\eqalign{ & \omega = 2\pi \nu \Rightarrow \nu = \frac{\omega }{{2\pi }} \cr & \therefore \nu = \frac{{qB}}{{2\pi m}} \cr} $$
NOTE
In the resultant expression $$\frac{q}{m}$$ is known as specific charge. It is sometimes denoted by $$\alpha .$$ So, in terms of $$\alpha ,$$ the above formula can be written as
$$\omega = B\alpha \,\,{\text{and}}\,\,\nu = \frac{{B\alpha }}{{2\pi }}$$

As, $$I = \frac{q}{t}.$$  So, for an electron revolving in a circular orbit of radius $$r$$

$$q = e\,\,{\text{and}}\,\,t = T$$
$$ \Rightarrow I = \frac{e}{T} = \frac{e}{{\frac{{2\pi }}{\omega }}} = \frac{{\omega e}}{{2\pi }} = \frac{{2\pi ne}}{{2\pi }} = ne$$
The magnetic field produced at the centre is
$$B = \frac{{{\mu _0}I}}{{2R}} = \frac{{{\mu _0}ne}}{{2r}}$$

$${B_L} = \frac{{{\mu _0}I}}{{2\pi R}}$$
If the radius is $$\frac{R}{n};$$  the number of turns will be $$n.$$
$$\eqalign{ & {B_C} = \frac{{n{\mu _0}I}}{{2\pi \left( {\frac{R}{n}} \right)}} = {n^2}\frac{{{\mu _0}I}}{{2\pi R}} \cr & {\text{Hence,}}\,\,\frac{{{B_L}}}{{{B_C}}} = \frac{1}{{{n^2}}} \cr} $$

The situation is shown in the figure.
$${F_E}$$ = Force due to electric field
$${F_B}$$ = Force due to magnetic field
It is given that the charged particle remains moving along $$X$$-axis (i.e. undeviated). Therefore $${F_B} = {F_E}$$
$$ \Rightarrow qvB = qE \Rightarrow B = \frac{E}{v} = \frac{{{{10}^4}}}{{10}} = {10^3}\,weber/{m^2}$$

Magnetic force between parallel wires per unit length is given by $$\frac{F}{l} = \frac{{{\mu _0}}}{{2\pi }} \times \frac{{{i_1}{i_2}}}{r}$$
where, $${{i_1}}$$ and $${{i_2}}$$ are the currents in wires 1 and 2 respectively and $$r$$ is the distance between them.
Since, it is given that between two wires there is a force of attraction, so, the direction of currents in both will be the same.
$$\eqalign{ & {\text{Given,}}\,{i_1} = {i_2} = 1A,r = 1\,m, \cr & {\mu _0} = 4\pi \times {10^{ - 7}}T{\text{ - }}m/A \cr & \therefore \frac{F}{l} = \frac{{4\pi \times {{10}^{ - 7}}}}{{2\pi }} \times \frac{{1 \times 1}}{1} \cr & = 2 \times {10^{ - 7}}N/m \cr} $$
NOTE
When current is in same direction in both the wires there will be attraction and if current in opposite direction there is repulsion.