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

Magnetic field B = 1 T
Mass of the proton, m_p = 1.67 x 10⁻²⁷kg
Charge of the proton, q = 1.60 x 10⁻¹⁹C
\(f=\frac { qB }{ { { 2\pi m }_{ p } } } =\frac { \left( 1.60\times { 10 }^{ -19 } \right) \left( 1 \right) }{ 2\left( 3.14 \right) \left( 1.67\times { 10 }^{ -27 } \right) } \)
= 15.3 x 10⁶ Hz = 15.3 MHz

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
\({ t }_{ 235 }=\frac { \text{ magnitude of the displacement} }{ velocity } \)
\(=\frac { 97.6\times { 10 }^{ -2 } }{ 1.00\times { 10 }^{ 5 } } =9.76\times { 10 }^{ -6 }s=9.76\mu s\)
\({ t }_{ 238 }=\frac { \text{magnitude of the displacement }}{ velocity } \)
\(=\frac { 98.8\times { 10 }^{ -2 } }{ 1.00\times { 10 }^{ 5 } } =9.88\times { 10 }^{ -6 }s=9.88\mu s\) 

When the current is passed through the loop, the torque is produced. If the torque acting on the loop is increased then the loop will start to rotate. The loop will start to lift if and only if the magnitude of magnetic torque due to current applied equals to the gravitational torque as shown in Figure
T_magnetic = T_{gravitational}
IAB = mgR
\(But\quad { p }_{ m }=IA=I(\pi { R }^{ 2 })\)
\(IR^{ 2 }B=mgR\)
\(\Rightarrow \frac { mg }{ \pi { R }B } \)
The current estimate using this equation should be applied so that one edge of loop is lifted off the surface.