KEY CONCEPT : We know that the magnetic field produced by a current carrying circular coil of radius $$r$$ at its centre is $$B = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{r} \times 2\pi \,$$
$$\eqalign{ & {\text{Here}}\,{B_A} = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{R} \times 2\pi \,{\text{and}}\,{B_B} = \frac{{{\mu _0}}}{{4\pi }}\frac{{2I}}{{2R}} \times 2\pi \cr & \Rightarrow \frac{{{B_A}}}{{{B_B}}} = 1 \cr} $$

Force between two long conductor carrying current,
$$\eqalign{ & F = \frac{{{\mu _0}}}{{4\pi }}\frac{{2{I_1}{I_2}}}{d} \times \ell \cr & F' = - \frac{{{\mu _0}}}{{4\pi }}\frac{{2\left( {2{I_1}} \right){I_2}}}{{3d}}\ell \,\,\,\therefore \frac{{F'}}{F} = \frac{{ - 2}}{3} \cr} $$

The force experienced by a charged particle moving in space where both electric and magnetic fields exist is called Lorentz force.
Due to both electric and magnetic fields, the total force experienced by the charged particle will be given by,
$$\eqalign{ & F = {F_e} + {F_m} = qE + q\left( {v \times B} \right) \cr & = q\left( {E + v \times B} \right) \cr} $$
When $$v,E$$  and $$B$$ are mutually perpendicular to each other. In this situation, if $$E$$ and $$B$$ are such that $$F = {F_e} + {F_m} = 0,$$    then acceleration in the particle, $$a = \frac{F}{m} = 0.$$
It means particle will go undeflected.

Magnetic field due to a long straight conductor carrying current $$i$$ at a distance $$r$$ is given by
$$B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2i}}{r}$$
Given, $${r_1} = 4\,cm,\,{r_2} = 12\,cm.$$
As $$B \propto \frac{1}{r}$$
and distance becomes 3 times, field is reduced to its one-third value.
Hence, $$B' = \frac{B}{3} = \frac{{{{10}^{ - 3}}}}{3} = 3.33 \times {10^{ - 4}}T$$

Due to electric field, it experiences force and decelerates i.e. its velocity decreases.

Given : $${M_1} = 1.20A{m^2}\,{\text{and}}\,{M_2} = 1.00A{m^2}$$
$$\eqalign{ & r = \frac{{20}}{2}\;cm = 0.1\;m \cr & {B_{{\text{net}}}} = {B_1} + {B_2} + {B_H} \cr & {B_{{\text{net}}}} = \frac{{{\mu _0}}}{{4\pi }}\frac{{\left( {{M_1} + {M_2}} \right)}}{{{r^3}}} + {B_H} \cr & = \frac{{{{10}^{ - 7}}\left( {1.2 + 1} \right)}}{{{{\left( {0.1} \right)}^3}}} + 3.6 \times {10^{ - 5}} = 2.56 \times {10^{ - 4}}wb/{m^2} \cr} $$

The workdone, $$dW = Fds\cos \theta $$
The angle between force and displacement is 90°.
Therefore work done is zero.

Case of positively charged particle :
Two force are acting on the positively charged particle
(a) due to electric field in the positive $$x$$-direction.
(b) Force due to magnetic field.
$$\eqalign{ & \overrightarrow F = q\left( {\overrightarrow v \times \overrightarrow B } \right) \cr & \Rightarrow \overrightarrow F = q\left( {v\hat i \times B\hat k} \right) \Rightarrow \overrightarrow F = qvB\left( { - \hat j} \right) \cr} $$
This forces will move the positively charged particle towards Y-axis.
Case of negatively charged particle.
Two forces are acting on the negatively charged particle
(a) due to electric field in the negative $$X$$-direction.
(b) due to magnetic field
$$\eqalign{ & \overrightarrow F = - q\left( {\overrightarrow v \times \overrightarrow B } \right) \cr & \overrightarrow F = - q\left[ {v\left( { - \hat i} \right) \times B\left( {\hat k} \right)} \right] \cr & \overrightarrow F = - qvB\left[ {\hat i \times \hat k} \right],\overrightarrow F = qvB\left( { - \hat j} \right) \cr} $$
Same direction as that of positive charge.
(C) is the correct answer.

We know, $$\vec F = I\left( {\vec \ell \times \vec B} \right)\,\,{\text{or}}\,\,d\vec F = IBd\ell \sin \theta $$
$$\eqalign{ & \therefore F = \int_{{{60}^ \circ }}^{{{120}^ \circ }} {I{B_0}d\ell \,\sin \theta d\ell } \,\,{\text{where,}}\,d\ell = rd\theta \cr & F = \int_{{{60}^ \circ }}^{{{120}^ \circ }} {I{B_0}} R\sin \theta d\theta = IBR\left[ { - \cos \theta } \right]_{{{60}^ \circ }}^{{{120}^ \circ }} \cr & = I{B_0}R\left[ { - \left( { - \frac{1}{2}} \right) + \left( {\frac{1}{2}} \right)} \right] = I{B_0}R \cr} $$
Hence, force acting on the arc is $$I{B_0}R\,\hat k.$$

If $$E$$ is the electric field strength and $$B$$ is the magnetic field strength and $$q$$ is the charge on a particle, then electric force on the charge
$${F_e} = qE$$
and magnetic force on the charge
$${F_m} = q\left( {v \times B} \right)$$
The net force on the charge
$$F = {F_e} + {F_m} = qE + q\left( {v \times B} \right)$$
Alternative
According to Lorentz force if a charged particle is in both electric field $$\left( E \right)$$  and magnetic field $$\left( B \right),$$  force is given by
$$F = q\left[ {E + \left( {v \times B} \right)} \right]$$