The wires at $$A$$ and $$B$$ are perpendicular to the plane of paper and current points is towards the reader. Let us consider certain points.
Point C (mid point between $$A$$ and $$B$$ ) : The magnetic field at $$C$$ due to $$A\left( {{{\overrightarrow B }_{CA}}} \right)$$   is in upward direction but magnetic field at $$C$$ due to $$B$$ is in downward direction. Net field is zero.
Point E : Magnetic field due to $$A$$ is upward and and magnetic field due to $$B$$ is downward but $$\left| {{{\overrightarrow B }_{EA}}} \right| < \left| {{{\overrightarrow B }_{EB}}} \right|.$$
∴ Net magnetic field is in downward direction.
Point D: $$\left| {{{\overrightarrow B }_{DA}}} \right| > \left| {{{\overrightarrow B }_{DB}}} \right|$$     Net field upwards. Similarly, other points can be considered.

The magnetic field $$\left( B \right)$$ at the centre of circular current carrying coil of radius $$R$$ and current $$I$$ is $$B = \frac{{{\mu _0}I}}{{2R}}$$
Similarly, if current is $$2I,$$  then
Magnetic field $$ = \frac{{{\mu _0}2I}}{{2R}} = 2B$$
So, resultant magnetic field $$ = \sqrt {{B^2} + {{\left( {2B} \right)}^2}} = \sqrt {5{B^2}} $$
$$ = \sqrt 5 B = \frac{{{\mu _0}I\sqrt 5 }}{{2R}}$$

$$\eqalign{ & {\text{As}}\,\,r = \frac{{mv}}{{qB}} = \frac{P}{{qB}} \cr & \therefore {\text{Area,}}\,\,A = \pi {r^2} = \pi {\left( {\frac{P}{{qB}}} \right)^2} = \frac{{\pi {P^2}}}{{qB}} = \frac{{2m\pi }}{{qB}}K \cr} $$

Frequency, $$\nu = \frac{{eB}}{{2\pi m}}$$
$$KE = \frac{1}{2}m{\nu ^2}\,\,{\text{and}}\,\,{\text{radius}}\,R = \frac{{m\nu }}{{eB}}$$
Here, velocity, $$\nu = \frac{{\pi R}}{{\frac{T}{2}}} = \frac{{2\pi R}}{T} = 2\pi R\nu $$
$$\therefore {\text{Radius,}}\,R = \frac{{m\left( {2\pi R\nu } \right)}}{{eB}}$$
Magnetic field, $$B = \frac{{2\pi m\nu }}{e}$$
Kinetic energy, $$K = \frac{1}{2}m{\left( {2\pi R\nu } \right)^2}$$
$$ = 2m{\pi ^2}{\nu ^2}{R^2}$$

Time period of cyclotron is
$$\eqalign{ & T = \frac{1}{\upsilon } = \frac{{2\pi m}}{{eB}};B = \frac{{2\pi m}}{e}\upsilon ;R = \frac{{m\upsilon }}{{eB}} = \frac{p}{{eB}} \cr & \Rightarrow P = eBR = e \times \frac{{2\pi m\upsilon }}{e}R = 2\pi m\upsilon R \cr & K.E.{\text{ }} = \frac{{{p^2}}}{{2m}} = \frac{{{{(2\pi m\upsilon R)}^2}}}{{2m}} = 2{\pi ^2}m{\upsilon ^2}{R^2} \cr} $$

Field due to one side of loop at $$O = \frac{{{\mu _0}I}}{{4\pi \left( {\frac{a}{2}} \right)}}\left( {2\sin {{45}^ \circ }} \right)$$
Field at $$O$$ due to all four sides is along unit vector $${{\hat e}_z}$$
$$\therefore $$ Total field $$ = 4.\frac{{{\mu _0}I}}{{4\pi \left( {\frac{a}{2}} \right)}}\left( {2\sin {{45}^ \circ }} \right)$$
$$ = \frac{{2\sqrt 2 {\mu _0}I}}{{\pi a}}$$

Torque acting on equilateral triangle in a magnetic field $$B$$ is
$$\tau = M \times B,\tau = iAB\sin \theta \,\,\,\left[ {iA = M} \right]$$
Area of $$\Delta LMN$$
$$A = \frac{{\sqrt 3 }}{4}{l^2}\,\,{\text{and}}\,\,\theta = {90^ \circ }\,\,\left[ {l = {\text{sides of triangle}}} \right]$$
Substituting the given values in the expression for torque, we have

$$\tau = i \times \frac{{\sqrt 3 }}{4}{l^2}B\sin {90^ \circ } = \frac{{\sqrt 3 }}{4}i{l^2}B\,\,\left( {\because \sin {{90}^ \circ } = 1} \right)$$
Hence, $$l = 2{\left( {\frac{\tau }{{\sqrt 3 Bi}}} \right)^{\frac{1}{2}}}$$

NOTE : Magnetic lines of force form closed loops. Inside a magnet, these are directed from south to north pole.

KEY CONCEPT : The time period of a rectangular magnet oscillating in earth’s magnetic field is given by $$T = 2\pi \sqrt {\frac{I}{{\mu {B_H}}}} $$
where $$I$$ = Moment of inertia of the rectangular magnet
$$\mu $$ = Magnetic moment
$${{B_H}}$$ = Horizontal component of the earth’s magnetic field
Case 1 : $$T = 2\pi \sqrt {\frac{I}{{\mu {B_H}}}} \,\,{\text{where}}\,I = \frac{1}{{12}}M{\ell ^2}$$
Case 2 : Magnet is cut into two identical pieces such that each piece has half the original length. Then
$$\eqalign{ & T' = 2\pi \sqrt {\frac{{I'}}{{\mu '{B_H}}}} \cr & {\text{where }}I' = \frac{1}{{12}}\left( {\frac{M}{2}} \right){\left( {\frac{\ell }{2}} \right)^2} = \frac{I}{8}{\text{ and }}\mu ' = \frac{\mu }{2} \cr & \therefore \frac{{T'}}{T} = \sqrt {\frac{{I'}}{{\mu '}} \times \frac{\mu }{I}} = \sqrt {\frac{{\frac{I}{8}}}{{\frac{\mu }{2}}} \times \frac{\mu }{I}} = \sqrt {\frac{1}{4}} = \frac{1}{2} \cr} $$

Force on a charged particle in the magnetic field is
$$\left| F \right| = q\left| {v \times B} \right|\,\,{\text{or}}\,\,F = qvB\sin \theta $$
when angle between velocity $$v$$ and magnetic induction $$B$$ is $${180^ \circ }$$ or $${0^ \circ },$$  then
$$F = qvB\sin {180^ \circ } = 0\,\,\left[ {{\text{as}}\,\sin {{180}^ \circ }\,{\text{or}}\,\sin {0^ \circ } = 0} \right]$$

So, the charged particle remains unaffected.