KEY CONCEPT : The time period of a charged particle $$\left( {m,q} \right)$$  moving in a magnetic field (B) is $$T = \frac{{2\pi m}}{{qB}}.$$   The time period does not depend on the speed of the particle.

$$B$$ due to $$AB = \frac{{{\mu _0}}}{{4\pi }}.\frac{i}{d}\left[ {\sqrt 3 } \right]$$
$$\eqalign{ & = \frac{{{\mu _0}}}{{4\pi }}\frac{i}{{\frac{\ell }{2}}}\left[ {\sqrt 3 } \right] \cr & = \frac{{{\mu _0}i}}{{4\pi \ell }}2\sqrt 3 \cr} $$
Net magnetic field $$ = \frac{{{\mu _0}}}{{2\pi }}\frac{i}{\ell }\sqrt 3 \times 2$$
$$ = \frac{{{\mu _0}}}{\pi }\frac{i}{\ell }\sqrt 3 $$

$$\alpha = \frac{q}{m},$$   path of the particle will be a helix of time period,
$$T = \frac{{2\pi m}}{{{B_0}q}} = \frac{{2\pi }}{{{B_0}\alpha }}$$
The give time $$t = \frac{\pi }{{{B_0}\alpha }} = \frac{T}{2}$$
∴ Coordinates of particle at time $$t = \frac{T}{2}$$  would be $$\left( {vx\frac{T}{2},0, - 2r} \right)$$
Here, $$r = \frac{{m{v_0}}}{{{B_0}q}} = \frac{{{v_0}}}{{{B_0}\alpha }}$$
∴ The coordinate are $$\left( {\frac{{{v_0}\pi }}{{{B_0}\alpha }},0,\frac{{ - 2{v_0}}}{{{B_0}\alpha }}} \right)$$

Net magnetic field, $$B = \sqrt {B_1^2 + B_2^2} $$
$$\eqalign{ & = \sqrt {{{\left( {\frac{{{\mu _0}{I_1}}}{{2\pi d}}} \right)}^2} + {{\left( {\frac{{{\mu _0}{I_2}}}{{2\pi d}}} \right)}^2}} \cr & = \frac{{{\mu _0}}}{{2\pi d}}\sqrt {I_1^2 + I_2^2} . \cr} $$

$$\eqalign{ & {\text{For}}\,r < \frac{R}{2},\,\,\,\,B = 0 \cr & {\text{For}}\,\frac{R}{2} \leqslant r < R \cr & B = \frac{{{\mu _0}}}{2}\left[ {r - \frac{{{R^2}}}{{2r}}} \right]J \cr & {\text{For}}\,r > R,\,\,\,B = \frac{{{\mu _0}i}}{{2\pi r}} \cr} $$

The magnetic field in the different regions is given by

$$\eqalign{ & {B_1} = \frac{{{\mu _0}}}{{4\pi }} \times \frac{I}{R}\left( { - \hat k} \right) \cr & {B_3} = \frac{{{\mu _0}I}}{{4\pi R}}\left( { - \hat k} \right) \cr & {B_2} = \frac{{{\mu _0}I}}{{4\pi R}}\left( { - \hat i} \right) \cr} $$
The net magnetic field at the centre $$O$$ is
$$\eqalign{ & B = {B_1} + {B_2} + {B_3} \cr & = \frac{{{\mu _0}I}}{{4\pi R}}\left( { - 2\hat k} \right) + \frac{{{\mu _0}I}}{{4R}}\left( { - \hat i} \right) \cr & = - \frac{{{\mu _0}I}}{{4\pi R}}\left( {2\hat k + \pi \hat i} \right) \cr} $$

Magnetic field at the centre due to current in arc $$ABC$$  is
$${B_1} = \frac{{{\mu _0}}}{{4\pi }}\frac{{{I_1}}}{r}\theta \,\,\,\,\,\,\,\,\left( {{\text{Directed}}\,{\text{upwards}}} \right)$$
Magnetic field at the centre due to current in arc $$ADB$$  is
$${B_2} = \frac{{{\mu _0}}}{{4\pi }}\frac{{{I_2}}}{r}\left( {2\pi - \theta } \right)\,\,\,\,\,\,\left( {{\text{Directed}}\,{\text{downwards}}} \right)$$
Therefore net magnetic field at the centre
$$\eqalign{ & B = \frac{{{\mu _0}}}{{4\pi }}\frac{{{I_1}}}{r}\frac{\theta }{\pi } - \frac{{{\mu _0}}}{{4\pi }}\frac{{{I_2}}}{r}\left( {2\pi - \theta } \right) \cr & {\text{Also}},\,{I_1} = \frac{E}{{{R_1}}} = \frac{E}{{\frac{{\rho {\ell _1}}}{A}}} = \frac{{EA}}{{\rho r\theta }} \cr & {\text{and}}\,{I_2} = \frac{E}{{{R_2}}} = \frac{E}{{\frac{{\rho {\ell _2}}}{A}}} = \frac{{EA}}{{\rho r\left( {2\pi - \theta } \right)}} \cr & \therefore B = \frac{{{\mu _0}}}{{4\pi }}\left[ {\frac{{EA}}{{\rho r\theta }} \times \frac{\theta }{r} - \frac{{EA}}{{\rho r\left( {2\pi - \theta } \right)}} \times \frac{{\left( {2\pi - \theta } \right)}}{r}} \right] = 0 \cr} $$

Net current is $$\left( {20 - 6 + 12 - 7 + 18} \right)A,\,{\text{i}}{\text{.e}}{\text{.,}}\,37\,A$$
$$\eqalign{ & r = \frac{{10}}{{100}}m = \frac{1}{{10}}m \cr & B = \frac{{{\mu _0}I}}{{2\pi r}} = \frac{{4\pi \times {{10}^{ - 7}} \times 37 \times 10}}{{2\pi \times 1}} = 74 \times {10^{ - 6}}T \cr & = 74\,\mu T. \cr} $$

According to Lorentz force,
$${F_{{\text{net}}}} = q\left[ {E + \left( {v \times B} \right)} \right]$$
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 pass undeflected.
Here, $${F_e} = {F_m}$$
So, \[qE = qvB\,\,\left[ {\begin{array}{*{20}{c}} {{F_e} = qE}\\ {{F_M} = qvB} \end{array}} \right]\]
$$\eqalign{ & {\text{or}}\,\,v = \frac{E}{B} \cr & {\text{Given,}}\,\,E = 20\,V{m^{ - 1}} \cr & B = 0.5\,T \cr & \therefore v = \frac{{20}}{{0.5}} = 40\,m/s \cr} $$

Given : $$f = 24 \times {10^6}\,Hz,R = 0.60\,m$$
We know that, $$R = \frac{{mv}}{{qB}}$$
$$\eqalign{ & \therefore B = \frac{{mv}}{{qR}}\,......\left( {\text{i}} \right) \cr & {\text{where}}\,\,v = \omega R = 2\pi fR \cr & = 2\pi \times 24 \times {10^6} \times 0.60 \cr & = 9.04 \times {10^7}\,m/s \cr} $$
From (i),
$$B = \frac{{\left( {3.34 \times {{10}^{ - 27}}} \right)\left( {9.04 \times {{10}^7}} \right)}}{{1.6 \times {{10}^{ - 19}} \times 0.60}} = 3.2\,T.$$