Given, l = 50cm
Number of turns per metre = 1000
Total number of turns (N) = 1000 x 1/2 = 500
Area, A = 2 x 10^-4 m²
Currecnt, I = 2 A
Magnetic field, B = 0.32 T
Torque, = \(MB\sin { \theta } \quad \)
\(= (NIA)B\sin { \theta } \quad \)
= 500  x 2 x (2 x 10^-4) x 0.32 
\( \tau \) ​​​​​= 0.032 N - m

Given, magnetic length of bar magnet, 2l = 6 cm
l = 3 cm = 3 x 10^-2 m
Distance moment, M = 0.60 A-m²  \({ B }_{ 2 }=\frac { { \mu }_{ 0 } }{ 4\pi } \times \frac { M }{ \left( { d }^{ 2 }+{ l }^{ 2 } \right) ^{ 3/2 } } \)
 \(=\frac { { \mu }_{ 0 }M }{ 4\pi { d }^{ 3 } } \)
 \(=\frac { 4\pi \times { 10 }^{ -7 }\times 0.60 }{ 4\pi { \times \left( 0.6 \right) }^{ 3 } }\)
 B₂ = 2.7 x 10^-7T

Given, pole strength, m = 6.64 A-m          
r = 3 mm = 3 x 10^-3 m  [ m₁ = m₂ = m₃]
Since, force, F = \(F=\frac { { \mu }_{ 0 } }{ 4\pi } \times \frac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } \)  
\(F=\frac { { \mu }_{ 0 } }{ 4\pi } \times \frac { _{ { m }^{ 2 } } }{ { r }^{ 2 } } \)
\(=\frac { { \mu }_{ 0 } }{ 4\pi } \times \frac { { \left( 6.64 \right) }^{ 2 } }{ { \left( { 3\times }{ 10 }^{ -3 } \right) }^{ 2 } } \)
​\(=\frac { { 4\pi \times }{ 10 }^{ -7 } }{ 4\pi } \times \frac { 44.0896 }{ 9\times { 10 }^{ -6 } } ={ 10 }^{ -1 }\times 4.8988\)
 F = 0.49N

Given,  V = 10 MHz = 107 Hz
r₀ = 50 cm = 0.50 m
m_\(\alpha\) = 4.0028 x 1.66 x 10^-27kg
= 6.645 x 10^-27 kg
q = 2e.= 2 x 1.6 x10 ^-19 = 3.2 x 10^-19C
\(\because \ v=\frac{B q}{2 \pi m_{\alpha}}\)
\(\therefore \ B=\frac{2 \pi m_{\alpha} v}{q}=\frac{2 \times 22}{7} \times \frac{6.645 \times 10^{-27} \times 10^{7}}{3.2 \times 10^{-19}}\)
= 1.305 T
Maximum kinetic energy is given by
\(E_{\max }=\frac{B^{2} q^{2} r^{2}}{2 m_{\alpha}}=\frac{(1.305)^{2} \times(3.2)^{2} \times 10^{-38} \times 0.25}{2 \times 6.645 \times 10^{-27} \times 1.6 \times 10^{-13}} \mathrm{MeV}\)
= 20.5 MeV 

Given, current, (I)= 10 A (East to West)
Distance, (r) =10 cm = 10 x 10^-2 m
Magnetic field, \(|B|=?\) 

The magnitude of magnetic field \(|B|\) for infinite length of
\(wire=\frac { { \mu }_{ 0 }I }{ 2\pi r } \)
\(\Rightarrow \ |B|=\cfrac { 4\pi \times{ 10 }^{ -7 } \times 10 }{ 2\times \pi \times10\times { 10 }^{ -2 } } =2\times{ 10 }^{ -5 }T\)         
The direction of magnetic field is given by right hand thumb rule or Maxwell's cork screw rule. So, the direction of magnetic field at point 10 cm North due to flowing current is perpendicularly inwards to the plane of paper.