In a right angle triangle OPN let the angle \(\angle\)OPN = \(\theta \)₁ which implies, \({ \varphi }_{ 1 }=\frac { \pi }{ 2 } -{ \theta }_{ 1 }\) and also in a right angle triangle OPM, 
\(\angle\)OPN = \(\theta \)₂ which implies, \({ \varphi }_{ 2 }=\frac { \pi }{ 2 } +{ \theta }_{ 2 }\)
Hence,
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 4\pi a } \left( cos\left( \frac { \pi }{ 2 } -{ \theta }_{ 1 } \right) -cos\left( \frac { \pi }{ 2 } +{ \theta }_{ 2 } \right) \right) \hat { n } \)
\(=\frac { { \mu }_{ ° }I }{ 4\pi a } (si{ n }_{ 1 }+{ sin }_{ 2 })\hat { n } \)

Given that I = 1 A and radius r = 1 m
B_straightwire =  \(=\frac { { \mu }_{ ° }I }{ 2\pi r } =\frac { 4\pi \times { 10 }^{ -7 }\times 1 }{ 2\pi \times 1 } =2\times { 10 }^{ -7 }T\)
But the Earth’s magnetic field is B_earth \(\sim { 10 }^{ -5 }T\)
So, B_straightwire is one hundred times smaller than B_Earth.

Velocity of the particle is \(\vec { v } =v\hat { j } \)
(a) Magnetic field is along positive y-direction, this implies \(\vec B=B\hat { j } \)

From Lorentz force, \( {\vec F } _{ m }=q(v\hat { j } \times B\hat { j } )=\vec 0\)
So, no force acts on the particle when it moves along the direction of magnetic field.
(b) Since the magnetic field points in positive z - direction, this implies, \(\vec { B } =B\hat { k } \)

From Lorentz force, \( {\vec F } _{ m }=q(v\hat { j } \times B\hat { k } )=qvB\vec i \)
Therefore, the magnitude of the Lorentz force is qvB and direction is along positive x - direction.
(c) Magnetic field is in zy - plane and making an angle θ with the velocity of the particle, which implies \( {\vec B } =Bcos\theta \hat { j } +Bsin\theta \hat { k } \)

From Lorentz force,
\({ \vec F }_{ m }=q(v\hat { j } )\times (Bcos\theta \hat { j } +Bsin\theta \hat k)\) 
\(=qvBsin\theta \hat { i } \)

Charge of an electron q = –1.60 × 10^–19 C ⇒ |q| = 1 60 x 10^-19 C
Magnitude of magnetic field B = 0.500 T
Mass of the electron, m = 9.11 × 10^–31 kg
Radius of the orbit, r = 2.50 mm = 2.50 × 10^–3 m
Speed of the electron,  V = \(q \frac{\mathrm{rB}}{\mathrm{m}}\)
\( v = 1.60 \times 10^{-19} \times\frac{ 2.50 \times 10^{-3} \times 0.500}{9.11 \times 10^{-31}}\)
\(v=2.195 \times 10^8 \mathrm{~m} \mathrm{s} ^{-1}\)

Since isotopes are singly ionized, they have equal charge which is equal to the charge of an electron, q = - 1.6 x 10^-19 C. Mass of uranium \(_{ 92 }^{ 235 }{ U and _{ 92 }^{ 238 }{ U } }\) are 3.90 x 10^-25 kg and 3.95 x 10^-25 kg respectively. Magnetic field applied, B = 0.500 T. Velocity of the electron is 1.00 x 10⁵ m s^-1, then 
(a) the radius of the path of \(_{ 92 }^{ 235 }{ U }\) is r₂₃₅
\({ r }_{ 235 }=\frac { { m }_{ 235 }v }{ \left| q \right| B } =\frac { 3.90\times { 10 }^{ -25 }\times 1.00\times { 10 }^{ 5 } }{ 1.6\times { 10 }^{ -19 }\times 0.500 } =48.8\times { 10 }^{ -2 }m\)
r₂₃₅ = 48.8cm 
The diameter of the semi-circle due to \(_{ 92 }^{ 235 }{ U\ \ is \ \ { d }_{ 235 }=2{ r }_{ 235 } }\) = 97.6 cm
The radius of the path of _{\(_{ 92 }^{ 238 }{ U\ is\ 2{ r }_{ 238 }\ then}\)}
\({ r }_{ 238 }=\frac { { m }_{ 238 }v }{ \left| q \right| B } =\frac { 3.90\times { 10 }^{ -25 }\times 1.00\times { 10 }^{ 5 } }{ 1.6\times { 10 }^{ -19 }\times 0.500 } =49.4\times { 10 }^{ -2 }m\)
r₂₃₈ = 49.4 cm
The diameter of the semi-circle due to  \(^{ 238 }_{92}{ U\ is \ 2{ r }_{ 238 } \ =98.8 \ cm}\)
Therefore the separation distance between the isotopes is \(\triangle d={ d }_{ 238 }-{ d }_{ 235 }=1.2 \ cm\)
(b) The time taken by each isotope to complete one semi-circular path are
\({ t }_{ 235 }=\frac { \text{ magnitude of the displacement} }{ velocity } \)
\(=\frac { 97.6\times { 10 }^{ -2 } }{ 1.00\times { 10 }^{ 5 } } =9.76\times { 10 }^{ -6 }s=9.76\mu s\)
\({ t }_{ 238 }=\frac { \text{magnitude of the displacement }}{ velocity } \)
\(=\frac { 98.8\times { 10 }^{ -2 } }{ 1.00\times { 10 }^{ 5 } } =9.88\times { 10 }^{ -6 }s=9.88\mu s\)