If there are $$N$$ turns in a coil, $$i$$ is the current flowing and $$A$$ is the area of the coil, then magnetic dipole moment or simply magnetic moment of the coil is $$M = NiA$$
As we know when velocity of charged particle entering to magnetic field region is perpendicular to $$B,$$ then it follows circular path.

As force on wire $$B$$ due to $$A$$ and $$C$$ are attractive, so we have following condition

$$F = \frac{{{\mu _0}{I^2}}}{{2\pi d}}$$
Resultant force on $$B$$
$$ = \sqrt {F_1^2 + F_2^2} = \sqrt 2 F = \sqrt 2 \times \frac{{{\mu _0}{I^2}}}{{2\pi d}} = \frac{{{\mu _0}{I^2}}}{{\sqrt 2 \pi d}}$$

As shown in the figure, the magnetic lines of force are directed from south to north inside a bar magnet.

Given : Current through the galvanometer,
$${i_g} = 5 \times {10^{ - 3}}A$$
Galvanometer resistance, $$G = 15\,\Omega $$
Let resistance $$R$$ to be put in series with the galvanometer to convert it into a voltmeter.
$$\eqalign{ & V = {i_g}\left( {R + G} \right) \cr & 10 = 5 \times {10^{ - 3}}\left( {R + 15} \right) \cr & \therefore R = 2000 - 15 = 1985 \cr & = 1.985 \times {10^3}\,\Omega \cr} $$

For loop $$B = \frac{{{\mu _0}nI}}{{2a}}$$
where, $$a$$ is the radius of loop.
Then, $${B_1} = \frac{{{\mu _0}I}}{{2a}}$$
Now, for coil $$B = \frac{{{\mu _0}I}}{{4\pi }}.\frac{{2nA}}{{{x^3}}}$$
at the centre $$x$$ = radius of loop
$$\eqalign{ & {B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2 \times 3 \times \left( {\frac{I}{3}} \right) \times \pi {{\left( {\frac{a}{3}} \right)}^2}}}{{{{\left( {\frac{a}{3}} \right)}^3}}} = \frac{{{\mu _0}.3I}}{{2a}} \cr & \therefore \frac{{{B_1}}}{{{B_2}}} = \frac{{\frac{{{\mu _0}I}}{{2a}}}}{{\frac{{{\mu _0}.3I}}{{2a}}}} \cr & {B_1}:{B_2} = 1:3 \cr} $$

At $$x = 2,{y^2} = 2x = 2 \times 2\,\,{\text{or}}\,\,y = 2.$$
Thus separation, $$AB = 4\,cm$$
The force, $$F = Bi\left( {AB} \right)$$
$$ = 4 \times 2 \times 4 = 32\,N.$$

Total magnetic field due to current carrying straight wire at any point $$P$$ is given by
$$B = \frac{{{\mu _0}i}}{{4\pi r}}\left( {\sin {\phi _1} + \sin {\phi _2}} \right)$$
When the conductor is of infinite length and point $$P$$ lies near the centre of conductor, then $${\phi _1} = {\phi _2} = {90^ \circ }$$
$$\eqalign{ & \therefore B = \frac{{{\mu _0}i}}{{4\pi r}}\left( {\sin {{90}^ \circ } + \sin {{90}^ \circ }} \right) \cr & = \frac{{{\mu _0}i}}{{4\pi r}} \cdot \frac{{2i}}{r} \Rightarrow r = \frac{{{\mu _0}i}}{{2\pi B}} \cr} $$
Here, current $$\left( i \right) = 12\,A,$$
magnetic field $$\left( B \right) = 3 \times {10^{ - 5}}\,Wb/{m^2}$$
$$\therefore $$ Perpendicular distance from wire to the point,
$$\eqalign{ & r = \frac{{4\pi \times {{10}^{ - 7}} \times 12}}{{2 \times \pi \times \left( {3 \times {{10}^{ - 5}}} \right)}} \cr & = 8 \times {10^{ - 2}}m \cr} $$

The magnetic field is $$B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2I}}{r}$$
$$\eqalign{ & = {10^{ - 7}} \times \frac{{2 \times 100}}{4} \cr & = 5 \times {10^{ - 6}}T \cr} $$

According to right hand palm rule, the magnetic field is directed towards south.

If both electric and magnetic fields are present and are perpendicular to each other and the particle is moving perpendicular to both of them with $${F_e} = {F_m}.$$   In this situation $$E \ne 0$$  and $$B \ne 0$$  and $${F_e} + {F_m} = 0.$$
But if electric field becomes zero, then only force due to magnetic field exists. And $$E$$ is perpendicular to the $$B$$ so the charge moves along a circle.

$$\eqalign{ & B = \frac{{{\mu _0}i{a^2}}}{{2{{\left( {{x^2} + {a^2}} \right)}^{\frac{3}{2}}}}} \cr & B' = \frac{{{\mu _0}i}}{{2a}} = \frac{{{\mu _0}i{a^2}}}{{2a{{\left( {{x^2} + {a^2}} \right)}^{\frac{3}{2}}}}}\left( {\frac{{{{\left( {{x^2} + {a^2}} \right)}^{\frac{3}{2}}}}}{{{a^2}}}} \right) \cr & B' = \frac{{B.{{\left( {{x^2} + {a^2}} \right)}^{\frac{3}{2}}}}}{{{a^3}}} \cr & {\text{Put}}\,\,x = 4\,\& \,a = 3 \Rightarrow B' = \frac{{54\left( {{5^3}} \right)}}{{3 \times 3 \times 3}} = 250\mu T \cr} $$