Motion of Charged Particle In Magnetic Field Questions in English

(B) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a semi-circular path of radius $R = \frac{mv}{qB}$.
Since the magnetic field is semi-infinite,the particles will complete a semi-circle and exit the field region,so $(A)$ is false.
Because the particles enter with the same velocity $v$ and the magnetic field $B$ is uniform,they will exit the field moving in the opposite direction to their entry,but still parallel to each other. Thus,$(B)$ is true.
The time taken to complete a semi-circle is $t = \frac{\pi m}{qB}$.
Since the mass of a proton $(m_p)$ is much greater than the mass of an electron $(m_e)$,the time taken for the proton to exit will be much greater than the time taken for the electron to exit. Thus,$(D)$ is true and $(C)$ is false.
Therefore,statements $(B)$ and $(D)$ are correct.

(D) The electric field $\vec{E}$ and magnetic field $\vec{B}$ are both along the $y$-axis. The Lorentz force is $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
$1$. If $\theta=0^{\circ}$,the initial velocity is $\vec{v} = v_0 \hat{i}$. The magnetic force $\vec{F}_B = q(v_0 \hat{i} \times B_0 \hat{j}) = q v_0 B_0 \hat{k}$ acts in the $z$-direction,causing circular motion in the $x-z$ plane. Simultaneously,the electric force $\vec{F}_E = q E_0 \hat{j}$ causes acceleration along the $y$-axis. Thus,the path is a helical motion with increasing pitch along the $y$-axis. Both $(A)$ and $(B)$ are incorrect.
$2$. If $\theta=10^{\circ}$,the velocity has components $v_x = v \cos \theta$ and $v_y = v \sin \theta$. The magnetic force depends only on $v_x$,causing circular motion in the $x-z$ plane. The electric force and $v_y$ cause accelerated motion along the $y$-axis. The combined motion is helical with increasing pitch. Thus,$(C)$ is correct.
$3$. If $\theta=90^{\circ}$,the velocity is $\vec{v} = v_0 \hat{j}$. Since $\vec{v}$ is parallel to $\vec{B}$,$\vec{F}_B = 0$. The particle only experiences the electric force $\vec{F}_E = q E_0 \hat{j}$,resulting in linear accelerated motion along the $y$-axis. Thus,$(D)$ is correct.
Therefore,the correct options are $(C)$ and $(D)$.

(A) The radius of the circular path of a charged particle in a magnetic field is given by $R = \frac{mv}{qB}$.
For the region $y > 0$,the radius is $R_1 = \frac{mv_0}{qB_1}$.
For the region $y < 0$,the radius is $R_2 = \frac{mv_0}{qB_2}$.
Given $B_2 = 4B_1$,we have $R_2 = \frac{mv_0}{q(4B_1)} = \frac{R_1}{4}$.
The particle moves in a semicircle of radius $R_1$ in the region $y > 0$ and then in a semicircle of radius $R_2$ in the region $y < 0$.
The total distance traveled along the $x$-axis is $\Delta x = 2R_1 + 2R_2 = 2R_1 + 2(\frac{R_1}{4}) = 2R_1 + \frac{R_1}{2} = \frac{5R_1}{2}$.
The time taken to complete a semicircle in region $y > 0$ is $t_1 = \frac{\pi m}{qB_1}$.
The time taken to complete a semicircle in region $y < 0$ is $t_2 = \frac{\pi m}{qB_2} = \frac{\pi m}{q(4B_1)} = \frac{t_1}{4}$.
The total time $T = t_1 + t_2 = t_1 + \frac{t_1}{4} = \frac{5t_1}{4}$.
The average speed along the $x$-axis is $v_{avg} = \frac{\Delta x}{T} = \frac{5R_1/2}{5t_1/4} = \frac{5R_1}{2} \times \frac{4}{5t_1} = 2 \frac{R_1}{t_1}$.
Substituting $R_1 = \frac{mv_0}{qB_1}$ and $t_1 = \frac{\pi m}{qB_1}$,we get $v_{avg} = 2 \frac{mv_0/qB_1}{\pi m/qB_1} = 2 \frac{v_0}{\pi}$.
Given $v_0 = \pi \text{ m s}^{-1}$,the average speed is $v_{avg} = 2 \frac{\pi}{\pi} = 2 \text{ m s}^{-1}$.

(B) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.
Since the particle is accelerated through a potential $V$,the kinetic energy $K = qV$.
Substituting this,we get $r = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.
For the $\alpha$-particle: $m_\alpha = 4 \text{ amu}$,$q_\alpha = 2e$.
For the sulfur ion: $m_s = 32 \text{ amu}$,$q_s = 1e$.
The ratio of the radii is $\frac{r_s}{r_\alpha} = \frac{\sqrt{m_s/q_s}}{\sqrt{m_\alpha/q_\alpha}} = \sqrt{\frac{m_s}{m_\alpha} \cdot \frac{q_\alpha}{q_s}}$.
Substituting the values: $\frac{r_s}{r_\alpha} = \sqrt{\frac{32}{4} \cdot \frac{2}{1}} = \sqrt{8 \cdot 2} = \sqrt{16} = 4$.

(C) The initial velocity is $\vec{u}_1 = 4\hat{i} \text{ m/s}$. The final velocity is $\vec{u}_2 = 2\sqrt{3}\hat{i} + 2\hat{j} \text{ m/s}$.
Since the magnetic force is perpendicular to the velocity,the speed remains constant: $|\vec{u}_1| = |\vec{u}_2| = 4 \text{ m/s}$.
The angle of deviation $\theta$ is given by $\tan \theta = \frac{v_y}{v_x} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$,so $\theta = 30^\circ = \frac{\pi}{6} \text{ rad}$.
The particle deflects in the $+y$ direction,implying the magnetic force $\vec{F} = Q(\vec{v} \times \vec{B})$ has a positive $y$-component. Given $\vec{v}$ is in the $x$-direction,$\hat{i} \times \vec{B}$ must have a positive $j$-component,which occurs if $\vec{B}$ is in the $-z$ direction.
The time taken to traverse the arc is $t = \frac{\theta}{\omega} = \frac{\theta M}{QB}$.
Given $t = 10 \text{ ms} = 0.01 \text{ s}$,we have $0.01 = \frac{\pi/6 \cdot M}{QB}$.
Solving for $B$: $B = \frac{\pi M}{6 \cdot 0.01 \cdot Q} = \frac{100\pi M}{6Q} = \frac{50\pi M}{3Q}$.
Thus,statements $(A)$ and $(C)$ are correct.

(D) The ion is accelerated through a potential $V$,so its kinetic energy is $K = qV = \frac{1}{2}mv^2$. The momentum $p = \sqrt{2mqV}$.
In the magnetic field,the ion follows a semi-circular path of radius $R = \frac{p}{qB} = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$.
The ion hits the detector at a distance $x = 2R = \frac{2}{B}\sqrt{\frac{2mV}{q}}$.
Given $m = A_M \times (5/3) \times 10^{-27} \text{ kg}$,$V = 192 \text{ V}$,$q = 1.6 \times 10^{-19} \text{ C}$,$B = 0.1 \text{ T}$.
$(A)$ For $H^{+}$ $(A_M = 1)$: $x = \frac{2}{0.1}\sqrt{\frac{2 \times (5/3) \times 10^{-27} \times 192}{1.6 \times 10^{-19}}} = 20 \times \sqrt{4 \times 10^{-8}} = 20 \times 2 \times 10^{-4} \text{ m} = 4 \times 10^{-3} \text{ m} = 0.4 \text{ cm}$. Wait,re-calculating: $x = 20 \times \sqrt{\frac{640}{1.6} \times 10^{-8}} = 20 \times \sqrt{400 \times 10^{-8}} = 20 \times 20 \times 10^{-4} = 0.04 \text{ m} = 4 \text{ cm}$. So $(A)$ is correct.
$(B)$ For $A_M = 144$: $x = 4 \text{ cm} \times \sqrt{144} = 4 \times 12 = 48 \text{ cm}$. So $(B)$ is correct.
$(C)$ For $A_M = 1$,$x_0 = 4 \text{ cm}$. For $A_M = 196$,$x_1 = 4 \text{ cm} \times \sqrt{196} = 4 \times 14 = 56 \text{ cm}$. The height is $x_1 - x_0 = 56 - 4 = 52 \text{ cm}$. So $(C)$ is correct.
$(D)$ For $A_M = 196$,$R = x/2 = 56/2 = 28 \text{ cm}$. So $(D)$ is incorrect.

(D) The magnetic force provides the centripetal force for the circular motion: $F_m = F_c \Rightarrow evB = \frac{mv^2}{r} \Rightarrow mv = eBr$.
According to Bohr's quantization condition for angular momentum: $mvr = \frac{nh}{2\pi}$.
Substituting $mv = eBr$ into the quantization condition: $(eBr)r = \frac{nh}{2\pi} \Rightarrow er^2B = \frac{nh}{2\pi}$.
Solving for $r$: $r = \sqrt{\frac{nh}{2\pi eB}}$.
For the first excited state,the principal quantum number is $n = 2$.
Substituting $n = 2$ into the expression: $r = \sqrt{\frac{2h}{2\pi eB}} = \sqrt{\frac{h}{\pi eB}}$.

(A) When the proton moves undeflected,the electric force equals the magnetic force: $eE = evB$,which gives $E = vB$ or $B = E/v$.
When the electric field is switched off,the proton moves in a circular path of radius $R$ due to the magnetic field: $R = \frac{mv}{eB}$.
Substituting $B = E/v$ into the radius formula: $R = \frac{mv}{e(E/v)} = \frac{mv^2}{eE}$.
Rearranging to solve for $E$: $E = \frac{mv^2}{eR}$.
Given: $m = 1.6 \times 10^{-27} \text{ kg}$,$v = 2 \times 10^5 \text{ ms}^{-1}$,$R = 2 \text{ cm} = 0.02 \text{ m}$,$e = 1.6 \times 10^{-19} \text{ C}$.
$E = \frac{(1.6 \times 10^{-27}) \times (2 \times 10^5)^2}{(1.6 \times 10^{-19}) \times 0.02} = \frac{1.6 \times 10^{-27} \times 4 \times 10^{10}}{1.6 \times 10^{-19} \times 2 \times 10^{-2}} = \frac{4 \times 10^{-17}}{2 \times 10^{-21}} = 2 \times 10^4 \text{ N/C}$.
Comparing this with $x \times 10^4 \text{ N/C}$,we get $x = 2$.

(B) The magnetic force acting on a charged particle moving in a magnetic field is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
For the electron to move in a straight line with constant velocity,the net magnetic force acting on it must be zero,i.e.,$\vec{F} = 0$.
Since $\vec{F} = qvB \sin \theta$,for $\vec{F} = 0$,the angle $\theta$ between the velocity vector $\vec{v}$ and the magnetic field vector $\vec{B}$ must be $0^\circ$ or $180^\circ$.
This implies that the magnetic field must be parallel or anti-parallel to the direction of the velocity of the electron.
Therefore,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(D) The magnetic field $\vec{B}$ due to the wire at a distance $r$ is $\vec{B} = \frac{\mu_0 I}{2 \pi r} (-\hat{k})$.
As the particle moves,the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ acts on it. Since the magnetic force is always perpendicular to the velocity,the speed $v_0$ remains constant.
Let the velocity be $\vec{v} = -v_x \hat{i} + v_y \hat{j}$. The force is $\vec{F} = q(-v_x \hat{i} + v_y \hat{j}) \times \frac{\mu_0 I}{2 \pi r} (-\hat{k}) = \frac{\mu_0 I q}{2 \pi r} (-v_x \hat{j} - v_y \hat{i})$.
The acceleration components are $a_x = -\frac{\mu_0 I q}{2 \pi m r} v_y$ and $a_y = -\frac{\mu_0 I q}{2 \pi m r} v_x$.
Using $\frac{v_x dv_x}{dr} = a_x = -\frac{\mu_0 I q}{2 \pi m r} v_y$ and $v_y = \sqrt{v_0^2 - v_x^2}$,we get $\int_{0}^{v_0} \frac{v_x dv_x}{\sqrt{v_0^2 - v_x^2}} = -\frac{\mu_0 I q}{2 \pi m} \int_{a}^{x_1} \frac{dr}{r}$.
Solving this gives $v_0 = \frac{\mu_0 I q}{2 \pi m} \ln(\frac{a}{x_1})$,so $x_1 = a e^{-\frac{2 \pi m v_0}{\mu_0 I q}}$.
Due to symmetry,the particle turns back at $x = x_1 e^{-\frac{2 \pi m v_0}{\mu_0 I q}} = a e^{-\frac{4 \pi m v_0}{\mu_0 I q}}$.

(C) The radius of curvature $r$ of a charged particle moving perpendicular to a magnetic field $B$ is given by $r = \frac{mv}{qB} = \frac{p}{qB}$,where $p$ is the momentum and $q$ is the magnitude of the charge.
For Assertion $A$: Both ions have the same momentum $p$. Thus,$r \propto \frac{1}{q}$. Since the charge of $O^{-2}$ is $2e$ and $H^{+}$ is $1e$,the radius of curvature for $O^{-2}$ is $r_{O} = \frac{p}{2eB}$ and for $H^{+}$ is $r_{H} = \frac{p}{eB}$. Since $r_{O} < r_{H}$,the curvature (which is $\frac{1}{r}$) of $O^{-2}$ is greater than that of $H^{+}$. Thus,Assertion $A$ is false.
For Reason $R$: For a proton and an electron with the same momentum $p$,the radius of curvature is $r = \frac{p}{qB}$. Since both have the same magnitude of charge $e$,their radii of curvature will be identical $(r_{p} = r_{e} = \frac{p}{eB})$. Thus,Reason $R$ is false.

(A) Given:
Charge $q = 1.6 \ \mu C = 1.6 \times 10^{-6} \ C$
Mass $m = 16 \ \mu g = 16 \times 10^{-9} \ kg$
Magnetic field $B = 6.28 \ T$
Since the particle is fired perpendicular to the magnetic field,it will perform uniform circular motion.
The time period $T$ for one complete revolution is given by the formula:
$T = \frac{2 \pi m}{qB}$
Substituting the values:
$T = \frac{2 \times 3.14 \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28}$
$T = \frac{6.28 \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28}$
$T = \frac{16 \times 10^{-9}}{1.6 \times 10^{-6}}$
$T = 10 \times 10^{-3} \ s = 0.01 \ s$
Thus,the time required to return to the original location is $0.01 \ s$.

(D) Since the velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,the charged particle moves in a circular path with radius $R = \frac{mv}{qB}$.
In region $2$,the particle completes a semi-circle of radius $R_2 = \frac{mv}{qB_2}$ and returns to the interface at point $D$. The distance $AD = 2R_2 = \frac{2mv}{qB_2}$.
Then,the particle enters region $1$ and completes a semi-circle of radius $R_1 = \frac{mv}{qB_1}$ and returns to the interface at point $C$. The distance $AC = 2R_1 = \frac{2mv}{qB_1}$.
The net displacement along the interface is the distance between the starting point $A$ and the final point $C$. Since the particle moves in opposite directions in the two regions relative to the interface,the displacement is $AC = AD - CD$ is incorrect based on the geometry. Looking at the path,the particle moves from $A$ to $D$ (in region $2$) and then from $D$ to $C$ (in region $1$).
The total displacement along the interface is $AC = AD - CD$ is not correct; rather,the particle moves from $A$ to $D$ and then from $D$ to $C$. The distance $AC = AD - CD$ is not the correct interpretation. The distance $AC = 2R_1 - 2R_2 = \frac{2mv}{qB_1} - \frac{2mv}{qB_2} = \frac{2mv}{q} \left(\frac{1}{B_1} - \frac{1}{B_2}\right) = \frac{2mv}{qB_1} \left(1 - \frac{B_1}{B_2}\right)$.

(D) The magnetic force provides the necessary centripetal force for circular motion:
$\frac{mv^2}{r} = qvB$
$mv = qBr$
$r = \frac{mv}{qB}$
Since kinetic energy $E = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2E}{m}}$.
Substituting $v$ in the expression for $r$:
$r = \frac{m}{qB} \sqrt{\frac{2E}{m}} = \frac{\sqrt{2mE}}{qB}$
Thus,$r = \left( \frac{\sqrt{2m}}{qB} \right) \sqrt{E}$
This implies $r \propto \sqrt{E}$.
The graph of $r$ versus $E$ is a parabola opening towards the $E$-axis,which corresponds to Graph $D$.

(B) The radius of a circular path for a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$,where $K$ is the kinetic energy.
From this,$K = \frac{r^2 q^2 B^2}{2m}$,which implies $K \propto \frac{q^2 r^2}{m}$.
For an $\alpha$-particle,$q_{\alpha} = 2e$ and $m_{\alpha} = 4m_p$. For a proton,$q_p = e$ and $m_p = m_p$.
Given $K_{\alpha} = 1 \text{ MeV}$ and $r_{\alpha} = r$,we want to find $K_p$ for $r_p = 2r$.
Using the ratio: $\frac{K_{\alpha}}{K_p} = \frac{q_{\alpha}^2 r_{\alpha}^2}{m_{\alpha}} \times \frac{m_p}{q_p^2 r_p^2}$.
Substituting the values: $\frac{1}{K_p} = \frac{(2e)^2 (r)^2}{4m_p} \times \frac{m_p}{e^2 (2r)^2} = \frac{4e^2 r^2}{4m_p} \times \frac{m_p}{4e^2 r^2} = \frac{1}{4}$.
Therefore,$K_p = 4 \text{ MeV}$.

(C) The velocity vector makes an angle of $30^{\circ}$ with the $y$-axis (the direction of the magnetic field $\vec{B}$). Since the velocity has a component parallel to the magnetic field,the path of the proton will be a helix.
The radius of the helical path is given by $r = \frac{mv_{\perp}}{qB} = \frac{mv \sin(30^{\circ})}{qB}$.
Substituting the values: $m = 1.67 \times 10^{-27} \ kg$,$v = 2 \times 10^6 \ m/s$,$\sin(30^{\circ}) = 0.5$,$q = 1.6 \times 10^{-19} \ C$,and $B = 0.104 \ T$.
$r = \frac{1.67 \times 10^{-27} \times 2 \times 10^6 \times 0.5}{1.6 \times 10^{-19} \times 0.104} \approx 0.1 \ m$.
The time period of the helical motion is $T = \frac{2\pi m}{qB}$.
$T = \frac{2 \times \pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.104} \approx 2\pi \times 10^{-7} \ s$.
Thus,the path is a helix of radius $0.1 \ m$ and time period $2\pi \times 10^{-7} \ s$.

(B) The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.
Since the proton is accelerated by a potential difference $V$,its kinetic energy is $K = qV = \frac{1}{2}mv^2$,which gives $v = \sqrt{\frac{2qV}{m}}$.
Substituting $v$ in the radius formula,we get $r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.
Given $V = 500 \times 10^3 \ V$,$B = 0.5 \ T$,$m = 1.6 \times 10^{-27} \ kg$,$q = 1.6 \times 10^{-19} \ C$,and $d = 0.1 \ m$.
$r = \frac{1}{0.5} \sqrt{\frac{2 \times 1.6 \times 10^{-27} \times 500 \times 10^3}{1.6 \times 10^{-19}}} = 2 \sqrt{\frac{2 \times 10^{-24} \times 500 \times 10^3}{10^{-19}}} = 2 \sqrt{1000 \times 10^{-2}} = 2 \sqrt{10} \approx 0.2 \ m$.
From the geometry of the path,$\sin \theta = \frac{d}{r} = \frac{0.1}{0.2} = 0.5$.
Therefore,$\theta = \arcsin(0.5) = 30^{\circ}$.

(B) The magnetic force on a charged particle moving with velocity $\overrightarrow{v}$ in a magnetic field $\overrightarrow{B}$ is given by the Lorentz force formula: $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
For an electron,the charge $q = -e$.
The velocity of the electron is $\overrightarrow{v} = v_x \hat{i}$ and the magnetic field is $\overrightarrow{B} = B_y \hat{j}$.
Substituting these into the force equation:
$\overrightarrow{F} = -e(v_x \hat{i} \times B_y \hat{j})$
$\overrightarrow{F} = -e v_x B_y (\hat{i} \times \hat{j})$
Since $\hat{i} \times \hat{j} = \hat{k}$,we get $\overrightarrow{F} = -e v_x B_y \hat{k}$.
The force is always perpendicular to both the velocity and the magnetic field. Since the velocity is in the $x$-direction and the magnetic field is in the $y$-direction,the force acts in the $z$-direction (specifically $-z$ for an electron). The particle will move in a circle in the plane perpendicular to the magnetic field,which is the $xz$-plane.

(C) For a particle of mass $m$ and charge $q$ moving in a magnetic field $B$,the radius of the circular path is given by $r = \frac{mv}{qB}$.
According to Bohr's quantization condition,the angular momentum is $mvr = \frac{nh}{2\pi}$.
Substituting $r = \frac{mv}{qB}$ into the quantization condition: $mv \left( \frac{mv}{qB} \right) = \frac{nh}{2\pi}$.
This simplifies to $\frac{m^2 v^2}{qB} = \frac{nh}{2\pi}$,which gives $mv^2 = \frac{nhqB}{2\pi m}$.
The kinetic energy $E$ is given by $E = \frac{1}{2} mv^2$.
Substituting the expression for $mv^2$: $E = \frac{1}{2} \left( \frac{nhqB}{2\pi m} \right) = \frac{nhqB}{4\pi m}$.

(A) According to the Bohr quantization condition,the angular momentum is given by $mvr = \frac{nh}{2 \pi}$.
Therefore,$vr = \frac{nh}{2 \pi m} \dots (i)$.
For a particle moving in a circular path under a magnetic field,the magnetic force provides the centripetal force: $qvB = \frac{mv^2}{r}$.
This simplifies to $mv = qBr$,or $v = \frac{qBr}{m} \dots (ii)$.
Substituting $v$ from $(ii)$ into $(i)$,we get $(\frac{qBr}{m})r = \frac{nh}{2 \pi m}$,which implies $r^2 = \frac{nh}{2 \pi qB}$.
The kinetic energy $E$ is given by $E = \frac{1}{2}mv^2$. Since $mv = qBr$,we have $E = \frac{1}{2}m(\frac{qBr}{m})^2 = \frac{q^2 B^2 r^2}{2m}$.
Substituting $r^2$ into the energy expression: $E = \frac{q^2 B^2}{2m} \times \frac{nh}{2 \pi qB} = \frac{nhqB}{4 \pi m}$.

(D) The magnetic force $F$ on a moving charge $q$ in a magnetic field $B$ is given by the Lorentz force formula: $F = q(v \times B)$.
Protons,cathode rays (electrons),and alpha particles are all charged particles. Therefore,they experience a magnetic force when moving through a magnetic field and are deflected.
Neutrons are electrically neutral particles,meaning their charge $q = 0$.
Since the charge is zero,the magnetic force $F = 0 \times (v \times B) = 0$.
Thus,neutrons cannot be deflected by a magnetic field.

(B) The kinetic energy gained by the electron accelerated through a potential difference $V$ is given by:
$KE = eV = \frac{1}{2} mv^2$
From this,the velocity $v$ of the electron is:
$v = \sqrt{\frac{2 eV}{m}}$
When a charged particle enters a magnetic field $B$ perpendicular to its velocity,it follows a circular path with radius $R$ given by:
$R = \frac{mv}{eB}$
Substituting the value of $v$ into the expression for $R$:
$R = \frac{m}{eB} \sqrt{\frac{2 eV}{m}} = \frac{1}{eB} \sqrt{m^2 \cdot \frac{2 eV}{m}} = \frac{1}{eB} \sqrt{2 Vme} = \sqrt{\frac{2 Vme}{e^2 B^2}} = \sqrt{\frac{2 Vm}{eB^2}}$

(A) The magnetic force $\vec{F}$ on a charged particle moving with velocity $\vec{V}$ in a magnetic field $\vec{B}$ is given by the Lorentz force formula: $\vec{F} = q(\vec{V} \times \vec{B})$.
Given $\vec{V} = a \hat{i}$ and $\vec{B} = b \hat{j} + c \hat{k}$.
Calculating the cross product: $\vec{V} \times \vec{B} = (a \hat{i}) \times (b \hat{j} + c \hat{k}) = ab(\hat{i} \times \hat{j}) + ac(\hat{i} \times \hat{k})$.
Using unit vector cross products $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{i} \times \hat{k} = -\hat{j}$,we get: $\vec{V} \times \vec{B} = ab \hat{k} - ac \hat{j}$.
Therefore,$\vec{F} = q(ab \hat{k} - ac \hat{j})$.
The magnitude of the force is $|\vec{F}| = q \sqrt{(ab)^2 + (-ac)^2} = q \sqrt{a^2b^2 + a^2c^2} = qa \sqrt{b^2 + c^2}$.

(C) The magnetic force $F$ acting on a charged particle moving with velocity $v$ in a magnetic field $B$ is given by the formula: $F = qvB \sin \theta$,where $\theta$ is the angle between the velocity vector and the magnetic field vector.
Since the charged particle is moving along a magnetic field line,the angle between its velocity and the magnetic field is $\theta = 0^{\circ}$.
Substituting this value into the force equation: $F = qvB \sin 0^{\circ}$.
Since $\sin 0^{\circ} = 0$,the magnetic force $F = 0$.

(D) The magnetic Lorentz force provides the necessary centripetal force for circular motion: $F = qvB = \frac{mv^2}{R}$.
From this,the radius $R$ is given by $R = \frac{mv}{qB}$.
Since the kinetic energy $K = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mK}$.
Substituting this into the radius formula: $R = \frac{\sqrt{2mK}}{qB}$.
This shows that $R \propto \sqrt{K}$.
If the energy is doubled $(K' = 2K)$,the new radius $R'$ is:
$\frac{R'}{R} = \sqrt{\frac{K'}{K}} = \sqrt{\frac{2K}{K}} = \sqrt{2}$.
Therefore,$R' = \sqrt{2}R$.

(A) When a charged particle enters a uniform magnetic field perpendicularly,it follows a circular path.
The radius of this path is given by:
$R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$
where $K$ is the kinetic energy of the particle.
Since the particle is accelerated through a potential difference $V$,its kinetic energy is $K = qV$.
Substituting this into the radius formula:
$R = \frac{\sqrt{2m(qV)}}{qB} = \frac{\sqrt{2mqV}}{qB}$
Squaring both sides:
$R^2 = \frac{2mqV}{q^2 B^2} = \frac{2mV}{qB^2}$
Given that the radius is $r$,we have $r^2 = \frac{2mV}{qB^2}$.
Solving for mass $m$:
$m = \frac{r^2 q B^2}{2V}$

(B) For a charged particle moving in a uniform magnetic field,the magnetic Lorentz force provides the necessary centripetal force for circular motion.
$qvB = \frac{mv^2}{R}$
$\therefore R = \frac{mv}{qB}$ ... $(i)$
Given the new speed $v' = \frac{v}{2}$ and the new magnetic field $B' = 2B$.
The new radius $R'$ is given by:
$R' = \frac{mv'}{qB'} = \frac{m(v/2)}{q(2B)}$
$R' = \frac{mv}{4qB}$
Substituting equation $(i)$ into the expression for $R'$:
$R' = \frac{R}{4}$

(C) The radius $r$ of a charged particle moving in a uniform magnetic field is given by the formula $r = \frac{mv}{qB}$.
Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mK}$.
Substituting this into the radius formula,we get $r = \frac{\sqrt{2mK}}{qB}$.
This shows that $r \propto \sqrt{K}$.
Let the initial kinetic energy be $K_1 = K$ and the initial radius be $R_1 = R$.
Let the new kinetic energy be $K_2 = 3K$ and the new radius be $R_2$.
Using the proportionality $r \propto \sqrt{K}$,we have $\frac{R_2}{R_1} = \sqrt{\frac{K_2}{K_1}}$.
Substituting the values,$\frac{R_2}{R} = \sqrt{\frac{3K}{K}} = \sqrt{3}$.
Therefore,the new radius is $R_2 = \sqrt{3} \cdot R$.

(D) When a charged particle moves in a perpendicular magnetic field, it undergoes uniform circular motion.
The magnetic force provides the necessary centripetal force: $\frac{mv^2}{R} = qvB$.
This simplifies to $v = \frac{qBR}{m}$.
The kinetic energy $E$ is given by $E = \frac{1}{2}mv^2 = \frac{1}{2}m \left(\frac{qBR}{m}\right)^2 = \frac{q^2B^2R^2}{2m}$.
Since $q$, $m$, and $R$ are constant, $E \propto B^2$.
If the magnetic field $B$ is increased by $4$ times $(B' = 4B)$, the new kinetic energy $E'$ will be:
$E' \propto (4B)^2 = 16B^2$.
Therefore, $E' = 16E$. The kinetic energy increases by $16$ times.

(C) The kinetic energy of the electrons is given by $\frac{1}{2}mv^2 = eV$,which implies the velocity $v = \sqrt{\frac{2eV}{m}}$.
The magnetic force experienced by the electrons is $F = evB = eB\sqrt{\frac{2eV}{m}}$.
This shows that $F \propto \sqrt{V}$.
Given that the new force $F^{\prime} = 2F$ when the potential is $V^{\prime}$,we have $\frac{F^{\prime}}{F} = \sqrt{\frac{V^{\prime}}{V}}$.
Substituting the values,$2 = \sqrt{\frac{V^{\prime}}{V}}$.
Squaring both sides,$4 = \frac{V^{\prime}}{V}$,which means $V^{\prime} = 4V$.
Therefore,the ratio $\frac{V}{V^{\prime}} = \frac{V}{4V} = \frac{1}{4}$.

(D) stationary electric charge produces an electric field in the space around it.
When an electric charge moves with a uniform velocity,it constitutes an electric current.
According to the Oersted experiment,an electric current produces a magnetic field in the surrounding space.
Since the charge is moving,it continues to possess its inherent electric field while simultaneously generating a magnetic field due to its motion.
Therefore,a charge moving with uniform velocity produces both electric and magnetic fields.

(B) The magnetic force acting on a charged particle is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Since the force is always perpendicular to the velocity vector $\vec{v}$,the work done by the magnetic force is $W = \int \vec{F} \cdot d\vec{r} = \int \vec{F} \cdot \vec{v} dt = 0$.
According to the work-energy theorem,the change in kinetic energy is equal to the work done,so the kinetic energy remains constant.
However,because the direction of the velocity vector changes continuously during the motion,the momentum $\vec{p} = m\vec{v}$ changes.
Therefore,the momentum changes while the kinetic energy remains constant.

(D) When a charged particle of mass $m$ and charge $q$ moves in a circular path perpendicular to a uniform magnetic field $B$,the magnetic Lorentz force provides the necessary centripetal force.
$qvB = \frac{mv^2}{r}$
From this,the radius of the path is $r = \frac{mv}{qB}$.
The time period $T$ is the time taken to complete one full circle,which is the circumference divided by the velocity:
$T = \frac{2 \pi r}{v} = \frac{2 \pi (mv/qB)}{v} = \frac{2 \pi m}{qB}$.
As seen from the formula $T = \frac{2 \pi m}{qB}$,the time period $T$ depends only on the mass $m$,charge $q$,and magnetic field $B$.
It is independent of the velocity $v$ of the charge.

(D) The magnetic force $F$ on a charged particle of mass $m$ and charge $q$ moving with velocity $v$ in a uniform magnetic field $B$ is given by $F = Bqv$.
When an electron is accelerated through a potential difference $V$,its kinetic energy is given by $\frac{1}{2}mv^2 = eV$.
From this,the velocity $v$ is $v = \sqrt{\frac{2eV}{m}}$.
Substituting this into the force equation,we get $F = B e \sqrt{\frac{2eV}{m}} = B e \sqrt{\frac{2e}{m}} \sqrt{V}$.
This shows that $F \propto \sqrt{V}$.
If the potential difference is increased to $2V$,the new force $F'$ will be $F' \propto \sqrt{2V}$.
Therefore,$\frac{F'}{F} = \frac{\sqrt{2V}}{\sqrt{V}} = \sqrt{2}$.
Thus,$F' = \sqrt{2}F$.

(B) The radius of curvature $r$ of a charged particle of mass $m$ and charge $q$ moving in a region with a uniform perpendicular magnetic field $B$ is given by the formula: $r = \frac{mv}{Bq} = \frac{p}{Bq}$.
Here,$p$ is the momentum of the charged particle.
Since the momenta $p$ are the same for both the electron and the proton,and the magnetic field $B$ is also the same,the radius of the circular path depends only on the charge $q$ of the particle.
However,the magnitude of the charge for both an electron and a proton is equal $(|q_e| = |q_p| = e)$.
Therefore,$r_e = \frac{p}{Be}$ and $r_p = \frac{p}{Be}$,which implies $r_e = r_p$.
Thus,both particles will follow a curved path of the same radius,neglecting the sense of revolution (direction of curvature).

(D) The magnetic field produced by a circular current-carrying loop at any point on its axis is directed along the axis of the loop.
Since the electron is projected along the same axis,its velocity vector $\vec{v}$ is parallel or anti-parallel to the magnetic field vector $\vec{B}$.
Therefore,the angle $\theta$ between the velocity and the magnetic field is either $0^{\circ}$ or $180^{\circ}$.
The magnetic force on a moving charge is given by the Lorentz force formula: $F = qvB \sin \theta$.
Substituting $\theta = 0^{\circ}$ or $180^{\circ}$,we get $\sin(0^{\circ}) = 0$ or $\sin(180^{\circ}) = 0$.
Thus,$F = 0$. The electron experiences no force.

(B) The magnetic force acting on a moving charge is given by $\overrightarrow{F} = q_0(\overrightarrow{v} \times \overrightarrow{B})$.
Since the force $\overrightarrow{F}$ is always perpendicular to the velocity vector $\overrightarrow{v}$,the work done by the magnetic force on the charge is zero $(W = \overrightarrow{F} \cdot \overrightarrow{d} = 0)$.
According to the work-energy theorem,the change in kinetic energy is equal to the work done,which implies that the kinetic energy remains constant.
Since kinetic energy $K = \frac{1}{2}mv^2$,a constant kinetic energy implies that the speed $v$ of the charge remains constant over time.
Therefore,the speed of $q_0$ after one second remains $v$.

(C) When a particle of charge $q$ and mass $M$ is accelerated through a potential difference $V$,its kinetic energy is given by $\frac{1}{2} M v^{2} = qV$.
Thus,the velocity $v = \sqrt{\frac{2qV}{M}}$.
When this particle enters a uniform magnetic field $B$ perpendicular to its velocity,it describes a circular path of radius $R = \frac{Mv}{qB}$.
Substituting the value of $v$,we get $R = \frac{M}{qB} \sqrt{\frac{2qV}{M}} = \frac{1}{B} \sqrt{\frac{2MV}{q}}$.
Since $q$,$V$,and $B$ are the same for both particles,$R \propto \sqrt{M}$,which implies $R^{2} \propto M$.
Therefore,$\frac{M_{A}}{M_{B}} = \left(\frac{R_{A}}{R_{B}}\right)^{2}$.

(D) The magnetic force provides the centripetal force for circular motion: $Bqv = \frac{mv^2}{R}$,which simplifies to $R = \frac{mv}{qB} = \frac{p}{qB}$,where $p$ is the momentum.
Since kinetic energy $K = \frac{p^2}{2m}$,we have $p = \sqrt{2mK}$.
Substituting this into the radius formula: $R = \frac{\sqrt{2mK}}{qB}$.
This shows that $R \propto \sqrt{K}$.
Given that the new kinetic energy $K_2 = 3K_1$,the new radius $R_2$ is related to the original radius $R_1$ by:
$\frac{R_2}{R_1} = \sqrt{\frac{K_2}{K_1}} = \sqrt{3}$.
Therefore,$R_2 = \sqrt{3} R$.

(B) The magnetic force $F$ acting on a charged particle moving with velocity $v$ in a magnetic field $B$ is given by the Lorentz force formula: $F = q(v \times B) = qvB \sin \theta$,where $\theta$ is the angle between the velocity vector and the magnetic field vector.
Since the particle is moving parallel to the magnetic field,the angle $\theta = 0^{\circ}$.
Substituting this into the formula: $F = qvB \sin(0^{\circ}) = qvB(0) = 0$.
Therefore,the magnetic force acting on the particle is zero.

(D) The kinetic energy $K$ of an electron accelerated by a potential $V$ is given by $K = eV = \frac{1}{2}mv^2$. Thus,the velocity $v$ is proportional to $\sqrt{V}$.
The magnetic force on a charged particle moving in a magnetic field is $F = qvB \sin \theta$. Assuming the angle $\theta$ remains constant,$F \propto v$.
Since $v \propto \sqrt{V}$,it follows that $F \propto \sqrt{V}$.
Given $F_{1} \propto \sqrt{V_{1}}$ and $F_{2} = 2F_{1} \propto \sqrt{V_{2}}$.
Dividing the two expressions: $\frac{F_{2}}{F_{1}} = \frac{\sqrt{V_{2}}}{\sqrt{V_{1}}} = 2$.
Squaring both sides: $\frac{V_{2}}{V_{1}} = 4$,which implies $\frac{V_{1}}{V_{2}} = \frac{1}{4}$.

(D) When a particle of charge $q$ and mass $m$ is accelerated through a potential difference $V$,its kinetic energy is given by $\frac{1}{2}mv^2 = qV$,which implies $v = \sqrt{\frac{2qV}{m}}$.
When this particle enters a uniform magnetic field $B$ perpendicular to its velocity,it follows a circular path of radius $r = \frac{mv}{qB}$.
Substituting the expression for $v$,we get $r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.
Since $q$,$V$,and $B$ are the same for both particles,$r \propto \sqrt{m}$,which means $r^2 \propto m$.
Therefore,the ratio of the masses is $\frac{m_X}{m_Y} = \left( \frac{r_1}{r_2} \right)^2$.

(A) The radius of a circular path for a charged particle in a perpendicular magnetic field is given by $r = \frac{mv}{qB}$.
For particle $a$,the radius is $r_{a} = \frac{m_{a} v_{a}}{q B}$.
For particle $b$,the radius is $r_{b} = \frac{m_{b} v_{b}}{q B}$.
Given that $r_{a} > r_{b}$,we substitute the expressions:
$\frac{m_{a} v_{a}}{q B} > \frac{m_{b} v_{b}}{q B}$.
Since the charge $q$ and magnetic field $B$ are the same for both particles,we can cancel them out.
Therefore,$m_{a} v_{a} > m_{b} v_{b}$.

(B) When an ion of charge $q$ and mass $m$ is accelerated through a potential difference $V$,its kinetic energy is given by: $E = qV = \frac{1}{2}mv^2$. From this,the velocity $v$ is $v = \sqrt{\frac{2qV}{m}}$.
When this ion enters a magnetic field $B$ perpendicular to its motion,it follows a circular path of radius $R$ due to the Lorentz force acting as the centripetal force: $qvB = \frac{mv^2}{R}$.
Substituting the expression for $v$ into the force equation: $qvB = \frac{m}{R} \left(\frac{2qV}{m}\right) = \frac{2qV}{R}$.
Simplifying for the charge-to-mass ratio: $qB = \frac{2qV}{vR} \implies B = \frac{mv}{qR} \implies R = \frac{mv}{qB}$.
Substituting $v = \sqrt{\frac{2qV}{m}}$ into the radius equation: $R = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.
Squaring both sides: $R^2 = \frac{2mV}{qB^2}$.
Rearranging to find the ratio $\frac{q}{m}$: $\frac{q}{m} = \frac{2V}{R^2 B^2}$.
Since $V$ and $B$ are constant,$\frac{q}{m} \propto \frac{1}{R^2}$.

(C) When a magnetic field is perpendicular to the motion of a charged particle,the magnetic force provides the necessary centripetal force.
$F_{c} = F_{m}$
$\frac{m v^{2}}{R} = B q v$
From this,the radius of the circular path is given by:
$R = \frac{m v}{B q}$
The time period $T$ of the circular motion is the time taken to complete one full circumference:
$T = \frac{2 \pi R}{v}$
Substituting the expression for $R$:
$T = \frac{2 \pi}{v} \left( \frac{m v}{B q} \right)$
$T = \frac{2 \pi m}{B q}$
Since $T$ depends only on the mass $m$,charge $q$,and magnetic field $B$,it is independent of both the radius $R$ and the speed $v$.