Motion of Charged Particle In Magnetic Field Questions in English

(C) When a charged particle enters a uniform magnetic field with a velocity that makes an acute angle $\theta$ (where $\theta \neq 0^\circ, 90^\circ, 180^\circ$) with the magnetic field,the velocity can be resolved into two components:
$1$. $A$ component parallel to the magnetic field $(v_{\parallel} = v \cos \theta)$,which causes the particle to move linearly along the field lines.
$2$. $A$ component perpendicular to the magnetic field $(v_{\perp} = v \sin \theta)$,which causes the particle to move in a circular path.
The combination of these two motions results in a helical path. Since the magnetic field is uniform and the angle $\theta$ remains constant,the pitch of the helix remains uniform.

(D) The electron has a velocity $\vec{v} = v \hat{i}$. For it to move in an anticlockwise circular path in the $xy$-plane,the initial magnetic force $\vec{F}_m$ must be directed along the positive $y$-axis (towards the center of the circular path).
Using the Lorentz force formula $\vec{F}_m = q(\vec{v} \times \vec{B})$,where $q = -e$ (for an electron),we have:
$\vec{F}_m = -e(v \hat{i} \times \vec{B}) = F_0 \hat{j}$.
This implies $(\hat{i} \times \vec{B}) = -\frac{F_0}{ev} \hat{j}$.
Since $\hat{i} \times \hat{k} = \hat{j}$,we have $\hat{i} \times (-\hat{k}) = -\hat{j}$.
Therefore,the magnetic field $\vec{B}$ must be directed along the negative $z$-axis.

(A) Given that the kinetic energy $K$ is the same for all particles: ${K_p} = {K_d} = {K_\alpha} = K$.
We know the mass and charge relations: $m_p = m$,$m_d = 2m$,${m_\alpha} = 4m$ and $q_p = e$,$q_d = e$,${q_\alpha} = 2e$.
The radius of a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.
For the proton: ${r_p} = \frac{\sqrt{2mK}}{eB}$.
For the deuteron: ${r_d} = \frac{\sqrt{2(2m)K}}{eB} = \sqrt{2} \left( \frac{\sqrt{2mK}}{eB} \right) = \sqrt{2} {r_p}$.
For the $\alpha$-particle: ${r_\alpha} = \frac{\sqrt{2(4m)K}}{(2e)B} = \frac{2\sqrt{2mK}}{2eB} = \frac{\sqrt{2mK}}{eB} = {r_p}$.
Comparing these,we get ${r_\alpha} = {r_p} < {r_d}$.

(D) When a charged particle of mass $m$ and charge $q$ moves with velocity $v$ perpendicular to a uniform magnetic field $B$,it experiences a magnetic Lorentz force $F = qvB$ which acts as the centripetal force.
Thus,$qvB = \frac{mv^2}{r}$,where $r$ is the radius of the circular path.
Solving for $r$,we get $r = \frac{mv}{qB}$.
The time period $T$ is the time taken to complete one circular orbit,given by $T = \frac{2\pi r}{v}$.
Substituting the value of $r$,we get $T = \frac{2\pi (mv/qB)}{v} = \frac{2\pi m}{qB}$.
From this expression,it is clear that 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 particle.

(B) The magnetic force on a moving charge is given by $F = qvB \sin \theta$. Since the motion is upward and the field is horizontal (south to north),the angle $\theta = 90^\circ$,so $F = qvB$.
Given kinetic energy $K = 200\, MeV = 200 \times 10^6 \times 1.6 \times 10^{-19}\,J = 3.2 \times 10^{-11}\,J$.
Using $K = \frac{1}{2}mv^2$,we find velocity $v = \sqrt{\frac{2K}{m}}$.
Substituting this into the force equation: $F = qB \sqrt{\frac{2K}{m}}$.
$F = (1.6 \times 10^{-19}) \times 5 \times \sqrt{\frac{2 \times 3.2 \times 10^{-11}}{1.67 \times 10^{-27}}}$.
$F = 8 \times 10^{-19} \times \sqrt{3.83 \times 10^{16}} \approx 8 \times 10^{-19} \times 1.95 \times 10^8 \approx 1.56 \times 10^{-10}\,N$.
Rounding to the nearest option,$F = 1.6 \times 10^{-10}\,N$.

(A) The magnetic force on a moving charge is given by the Lorentz force formula: $\overrightarrow{F_m} = q(\overrightarrow{v} \times \overrightarrow{B})$.
Here,the velocity vector $\overrightarrow{v}$ is in the $+X$-direction,and the magnetic field $\overrightarrow{B}$ is in the $-X$-direction.
The angle $\theta$ between $\overrightarrow{v}$ and $\overrightarrow{B}$ is $180^\circ$.
Since the cross product $\overrightarrow{v} \times \overrightarrow{B} = vB \sin(180^\circ) = 0$,the magnetic force $\overrightarrow{F_m}$ is zero.
Therefore,the charge will continue to move in the $X$-direction with constant velocity and remain unaffected.

(C) The radius of the circular path of a charged particle moving perpendicular to a magnetic field $B$ is given by $r = \frac{mv}{qB}$.
Since momentum $p = mv$,we can rewrite the expression as $r = \frac{p}{qB}$.
Given that both the electron and the proton have the same momentum $p$ and are moving in the same magnetic field $B$,the radius depends only on the charge $q$.
However,the magnitude of the charge of an electron and a proton is the same $(|q_e| = |q_p| = e)$.
Therefore,the radius of the path for both particles is $r = \frac{p}{eB}$,which is identical for both.
Thus,both paths are equally curved.

(C) According to the Lorentz force formula,$\vec{F} = q(\vec{v} \times \vec{B})$.
Here,the velocity $\vec{v}$ is directed towards the east,and the magnetic field $\vec{B}$ is directed vertically upwards.
Using the right-hand rule for the cross product $\vec{v} \times \vec{B}$,the force $\vec{F}$ acts in the horizontal plane towards the north.
Since the magnetic force is always perpendicular to the velocity of the particle,it does no work on the particle $(W = \vec{F} \cdot \vec{d} = 0)$.
Consequently,the kinetic energy and the speed of the particle remain constant.
$A$ particle moving perpendicular to a uniform magnetic field follows a circular path with a constant speed.

(B) The radius $r$ of the circular path of a charged particle moving in a uniform magnetic field is given by the formula $r = \frac{mv}{qB}$.
Here,$m$ is the mass of the particle,$v$ is its velocity,$q$ is the charge,and $B$ is the magnetic field strength.
From the formula,it is clear that the radius $r$ is directly proportional to the velocity $v$ of the particle $(r \propto v)$.
Therefore,if the velocity of the particle increases,the radius of the circular path also increases.

(A) When a charged particle moves in a uniform magnetic field,the magnetic force acting on it 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 field on the particle is $W = \int \vec{F} \cdot d\vec{s} = \int \vec{F} \cdot \vec{v} dt = 0$.
Because the work done is zero,the kinetic energy (and thus the total energy) of the particle remains constant.
However,the direction of the velocity vector changes continuously due to the magnetic force,which implies that the momentum $\vec{p} = m\vec{v}$ changes because momentum is a vector quantity.
Therefore,the momentum changes while the total energy remains the same.

(A) The Lorentz force acting on a charged particle moving in a magnetic field is given by the formula $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
Here,the charge of an electron is $q = -e$.
Since the electron is moving in the direction of the magnetic field,the angle $\theta$ between the velocity vector $\overrightarrow{v}$ and the magnetic field vector $\overrightarrow{B}$ is $0^\circ$.
The magnitude of the force is given by $F = |q|vB \sin \theta$.
Substituting $\theta = 0^\circ$,we get $F = evB \sin(0^\circ) = evB(0) = 0$.
Therefore,the force on the electron is zero.

(B) The radius $r$ of the circular path of a charged particle moving perpendicular to a magnetic field is given by the formula: $r = \frac{mv}{qB}$.
We are given the specific charge $\frac{q}{m} = 10^{11} \ C/kg$,which implies $\frac{m}{q} = \frac{1}{10^{11}} \ kg/C$.
Given values are $v = 10^7 \ m/s$ and $B = 10^{-4} \ T$.
Substituting these values into the formula:
$r = \left(\frac{m}{q}\right) \times \frac{v}{B} = \left(\frac{1}{10^{11}}\right) \times \frac{10^7}{10^{-4}}$
$r = \frac{10^7}{10^{11} \times 10^{-4}} = \frac{10^7}{10^7} = 1 \ m$.
Therefore,the radius is $1 \ m$.

(A) The radius $r$ of a charged particle moving in a uniform magnetic field is given by the formula $r = \frac{mv}{qB}$.
Since the kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$,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 radius be $R$ with energy $K$,and the new radius be $R_2$ with energy $2K$.
Then,$\frac{R_2}{R} = \sqrt{\frac{2K}{K}} = \sqrt{2}$.
Therefore,$R_2 = R\sqrt{2}$.

(D) When a charged particle with charge $q$,mass $m$,and velocity $v$ moves in a uniform magnetic field $B$ perpendicular to its motion,it experiences a magnetic Lorentz force $F = qvB$.
This force provides the necessary centripetal force for circular motion: $qvB = \frac{mv^2}{r}$.
Rearranging for the radius $r$,we get $r = \frac{mv}{qB}$.
Since linear momentum $P = mv$,the expression becomes $r = \frac{P}{qB}$.
From this formula,it is clear that $r$ is directly proportional to the linear momentum $P$ and inversely proportional to the magnetic field $B$.
Therefore,both statements $(b)$ and $(c)$ are correct.

(B) The magnetic force $F$ on a charged particle moving in a magnetic field is given by the formula: $F = qvB \sin \theta$.
Given:
Charge of a proton $q = 1.6 \times 10^{-19} \, C$
Velocity $v = 2.5 \times 10^7 \, m/s$
Magnetic field $B = 2.5 \, T$
Angle $\theta = 30^o$
Substituting the values:
$F = (1.6 \times 10^{-19}) \times (2.5 \times 10^7) \times 2.5 \times \sin(30^o)$
$F = (1.6 \times 10^{-19}) \times (6.25 \times 10^7) \times 0.5$
$F = 10 \times 10^{-12} \times 0.5$
$F = 5 \times 10^{-12} \, N$

(B) The magnetic force $F$ acting on a charge $q$ moving with velocity $v$ in a magnetic field $B$ is given by the Lorentz force formula: $F = q(v \times B)$.
In terms of magnitude,this is expressed as $F = qvB \sin \theta$,where $\theta$ is the angle between the velocity vector and the magnetic field vector.
From this formula,it is clear that the force depends on the charge $(q)$,the velocity $(v)$,the magnetic field $(B)$,and the angle $(\theta)$.
It does not depend on the mass $(m)$ of the particle.
Therefore,the correct option is $B$.

(A) The force on a moving charge in a magnetic field is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
Here,the charge of an electron is negative $(q = -e)$.
The velocity $\vec{v}$ is in the east direction,and the magnetic field $\vec{B}$ is in the upward direction.
Using the right-hand rule for the cross product $\vec{v} \times \vec{B}$: the direction of $\vec{v} \times \vec{B}$ is towards the north.
Since the electron has a negative charge,the force $\vec{F} = -e(\vec{v} \times \vec{B})$ will be in the opposite direction of $\vec{v} \times \vec{B}$.
Therefore,the force on the electron is towards the south.
Alternatively,using Fleming's left-hand rule for a negative charge: point the index finger upward (magnetic field),the middle finger towards the west (opposite to the electron's velocity),and the thumb will point towards the south.

(A) The magnetic force $F$ on a moving charge is given by the formula $F = qvB \sin(\theta)$.
Given:
Charge $q = 1\,C$
Velocity $v = 10\,m/s$
Magnetic field $B = 0.5\,T$
Angle $\theta = 90^\circ$ (since the velocity is perpendicular to the field).
Substituting the values:
$F = 1 \times 10 \times 0.5 \times \sin(90^\circ)$
$F = 1 \times 10 \times 0.5 \times 1$
$F = 5\,N$.

(D) The radius $r$ of a charged particle moving in a uniform magnetic field is given by the formula $r = \frac{mv}{qB}$.
Given the initial state: $r_1 = \frac{mv}{qB}$.
For the final state,the speed is doubled $(v_2 = 2v)$ and the magnetic field is halved $(B_2 = B/2)$.
The new radius $r_2$ is given by $r_2 = \frac{m(2v)}{q(B/2)}$.
Simplifying this expression: $r_2 = 4 \times \frac{mv}{qB} = 4r_1$.
Therefore,the resulting path has a radius of $4r$.

(D) The magnetic force $F$ acting on a charged particle moving in a magnetic field is given by the Lorentz force formula: $F = q(\vec{v} \times \vec{B}) = qvB \sin(\theta)$.
Here,$\theta$ is the angle between the velocity vector $\vec{v}$ and the magnetic field vector $\vec{B}$.
Since the electron enters the magnetic field in the same direction as the magnetic field,the angle $\theta = 0^\circ$.
Therefore,the magnetic force $F = qvB \sin(0^\circ) = 0$.
Since the net force on the electron is zero,there is no change in its velocity (both magnitude and direction remain constant). Thus,the velocity of the electron remains unchanged.

(D) When a charged particle of mass $m$ and charge $q$ enters a magnetic field $B$ perpendicularly with a velocity $v$,it experiences a magnetic Lorentz force given by $F = q(v \times B)$.
Since the velocity is perpendicular to the magnetic field,the magnitude of the force is $F = qvB$.
This magnetic force acts as the centripetal force required for circular motion,so $F = mv^2/r$.
Equating the two expressions: $qvB = mv^2/r$.
Solving for the radius $r$,we get $r = mv/qB$.

(B) The force on a moving charge in a magnetic field is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
Here,the charge of the electron is negative $(q = -e)$.
The velocity $\vec{v}$ is towards the east,and the magnetic field $\vec{B}$ is towards the north.
Using the right-hand rule for the cross product $\vec{v} \times \vec{B}$,the direction of $\vec{v} \times \vec{B}$ is vertically upward (from east to north).
Since the electron has a negative charge,the force $\vec{F} = -e(\vec{v} \times \vec{B})$ will be in the opposite direction of $\vec{v} \times \vec{B}$.
Therefore,the force on the electron is directed vertically downward.

(A) The period of revolution $T$ of a charged particle moving in a circular orbit in a uniform magnetic field is given by the formula:
$T = \frac{2\pi m}{qB}$
Given:
Mass $m = 9.0 \times 10^{-31} \ kg$
Charge $q = 1.6 \times 10^{-19} \ C$
Magnetic field $B = 1.0 \times 10^{-4} \ Wb/m^2$
Substituting the values into the formula:
$T = \frac{2 \times 3.14 \times 9.0 \times 10^{-31}}{1.6 \times 10^{-19} \times 1.0 \times 10^{-4}}$
$T = \frac{56.52 \times 10^{-31}}{1.6 \times 10^{-23}}$
$T = 35.325 \times 10^{-8} \ s \approx 3.5 \times 10^{-7} \ s$
Thus,the correct option is $A$.

(D) The magnetic force $\overrightarrow{F}$ on a charged particle moving in a magnetic field is given by the Lorentz force formula: $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
This can be expressed in magnitude as $F = qvB \sin(\theta)$,where $\theta$ is the angle between the velocity vector $\overrightarrow{v}$ and the magnetic field vector $\overrightarrow{B}$.
In this problem,the electron enters the magnetic field in the same direction as the field,which means the angle $\theta = 0^\circ$.
Since $\sin(0^\circ) = 0$,the force $F = qvB \sin(0^\circ) = 0$.
Therefore,the force on the electron is zero.

(A) The force on a charged particle moving in combined electric and magnetic fields is given by the Lorentz force equation: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
Since the proton is projected with its velocity $\vec{v}$ parallel to the electric field $\vec{E}$ and the magnetic field $\vec{B}$ is in the same direction as $\vec{E}$,the velocity $\vec{v}$ is also parallel to $\vec{B}$.
The magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B}) = 0$ because the angle between $\vec{v}$ and $\vec{B}$ is $0^\circ$.
The electric force $\vec{F}_e = q\vec{E}$ acts on the proton in the direction of the electric field,which is the same as the direction of its initial velocity.
Since there is a constant force acting in the direction of motion,the proton will continue to move in the same direction with an increasing velocity.

(A) When a charged particle (electron) moves perpendicular to a uniform magnetic field $B$,it experiences a magnetic Lorentz force $F = qvB \sin(90^\circ) = evB$.
This magnetic force acts as the centripetal force required for circular motion,given by $F_c = mv^2 / r$.
Equating the two forces: $evB = mv^2 / r$.
Solving for the radius $r$,we get $r = mv / eB$ or $r = mv / Be$.

(B) When a charged particle moves in a uniform magnetic field perpendicular to its velocity,it follows a circular path.
The radius $r$ of this circular path is given by the formula $r = \frac{mv}{qB}$.
Since the charge $q$ and the magnetic field $B$ are the same for both particles,we have $r \propto mv$.
From the figure,it is clear that the radius of the trajectory of particle $A$ is greater than the radius of the trajectory of particle $B$,i.e.,$r_A > r_B$.
Therefore,$m_A v_A > m_B v_B$.

(B) The radius of the circular path of a charged particle moving in a uniform magnetic field is given by $r = \frac{p}{qB}$,where $p$ is the momentum,$q$ is the charge,and $B$ is the magnetic field strength.
Since the radii $r$ and the magnetic field $B$ are the same for both particles,the momentum $p$ is directly proportional to the charge $q$ $(p \propto q)$.
Therefore,the ratio of the momentum of the proton $(p_p)$ to the alpha particle $(p_\alpha)$ is $\frac{p_p}{p_\alpha} = \frac{q_p}{q_\alpha}$.
We know that the charge of a proton is $q_p = e$ and the charge of an alpha particle is $q_\alpha = 2e$.
Substituting these values,we get $\frac{p_p}{p_\alpha} = \frac{e}{2e} = \frac{1}{2} = 0.5$.

(C) For the particle to move with a uniform velocity,the net force acting on it must be zero. This means the magnetic force must balance the gravitational force acting on the particle.
$|F_m| = mg$
Since the magnetic force is given by $F_m = qvB$ (assuming the magnetic field is perpendicular to the velocity),we have:
$qvB = mg$
$B = \frac{mg}{qv}$
Given:
$m = 0.6\, g = 0.6 \times 10^{-3}\, kg$
$q = 25\, nC = 25 \times 10^{-9}\, C$
$v = 1.2 \times 10^4\, m/s$
$g = 10\, m/s^2$
Substituting the values:
$B = \frac{0.6 \times 10^{-3} \times 10}{25 \times 10^{-9} \times 1.2 \times 10^4}$
$B = \frac{6 \times 10^{-3}}{30 \times 10^{-5}}$
$B = \frac{6 \times 10^{-3}}{0.3 \times 10^{-3}} = \frac{6}{0.3} = 20\, T$
Thus,the magnetic induction is $20\, T$.

(C) The radius of the circular path of a charged particle moving in a perpendicular magnetic field is given by $r = \frac{mv}{qB}$.
Since the velocity $v$ and magnetic field $B$ are the same for both particles,the ratio of the radii is $\frac{r_{\alpha}}{r_p} = \frac{m_{\alpha}}{m_p} \times \frac{q_p}{q_{\alpha}}$.
For an $\alpha$-particle,the mass $m_{\alpha} = 4m_p$ and the charge $q_{\alpha} = 2q_p$.
Substituting these values,we get $\frac{r_{\alpha}}{r_p} = \frac{4m_p}{m_p} \times \frac{q_p}{2q_p} = 4 \times \frac{1}{2} = \frac{2}{1}$.
Therefore,the ratio is $2:1$.

(B) The magnetic force on a moving charge is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
When the magnetic field $\vec{B}$ is applied in the direction of motion (parallel to velocity $\vec{v}$),the angle between $\vec{v}$ and $\vec{B}$ is $0^\circ$ or $180^\circ$.
Since the cross product $\vec{v} \times \vec{B} = vB \sin(\theta)$,and $\sin(0^\circ) = \sin(180^\circ) = 0$,the magnetic force $\vec{F}$ becomes zero.
Therefore,the motion of the electron remains unaffected by a magnetic field applied parallel to its direction of motion.

(A) The magnetic force acting on a charged particle moving in a magnetic field is given by $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
Since the force $\overrightarrow{F}$ is always perpendicular to the velocity vector $\overrightarrow{v}$,the work done by the magnetic force on the particle is $W = \int \overrightarrow{F} \cdot d\overrightarrow{s} = \int \overrightarrow{F} \cdot \overrightarrow{v} dt = 0$.
According to the work-energy theorem,the change in kinetic energy is equal to the work done by the net force.
Since the work done is zero,the change in kinetic energy is zero,which means the kinetic energy remains constant.

(B) When a charged particle enters a uniform magnetic field $B$ with a velocity $v$ perpendicular to the field,the magnetic Lorentz force $F = q(v \times B)$ acts on it.
This force is always perpendicular to both the velocity and the magnetic field.
Since the force is perpendicular to the velocity,it does not change the speed of the particle but changes its direction continuously.
This constant perpendicular force acts as a centripetal force,causing the particle to move in a circular path.
Cathode rays consist of a stream of electrons (negatively charged particles),which follow the same principle of circular motion under a perpendicular magnetic field.

(A) The frequency of revolution $\nu$ of a charged particle moving in a magnetic field is given by the formula: $\nu = \frac{qB}{2\pi m}$.
Since the magnetic field $B$ is constant and the velocity is the same,the frequency $\nu$ is directly proportional to the charge-to-mass ratio: $\nu \propto \frac{q}{m}$.
To find the minimum frequency,we need to find the particle with the minimum $\frac{q}{m}$ ratio.
For $Li^+$: $q = +1e$,$m \approx 7u$,so $\frac{q}{m} \approx \frac{1}{7}$.
For Electron: $q = -1e$,$m \approx \frac{1}{1836}u$,so $\frac{q}{m} \approx 1836$.
For Proton: $q = +1e$,$m \approx 1u$,so $\frac{q}{m} = 1$.
For $He^+$: $q = +1e$,$m \approx 4u$,so $\frac{q}{m} = 0.25$.
Comparing the ratios,$Li^+$ has the smallest $\frac{q}{m}$ value. Therefore,$Li^+$ will have the minimum frequency of revolution.

(C) The radius of the circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.
Since kinetic energy $K = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2K}{m}}$.
Substituting this into the radius formula,we get $r = \frac{m}{qB} \sqrt{\frac{2K}{m}} = \frac{\sqrt{2mK}}{qB}$.
Given that $K$ and $B$ are constant for all ions,the radius $r \propto \frac{\sqrt{m}}{q}$.
For $He^+$ ions: $m_1 = 4 \ amu$,$q_1 = 1e$.
For $O^{2+}$ ions: $m_2 = 16 \ amu$,$q_2 = 2e$.
Calculating the ratio of radii: $\frac{r_{He^+}}{r_{O^{2+}}} = \sqrt{\frac{m_1}{m_2}} \times \frac{q_2}{q_1} = \sqrt{\frac{4}{16}} \times \frac{2}{1} = \frac{1}{2} \times 2 = 1$.
Since the radii of the paths are equal,all ions will be deflected equally.

(D) The radius $r$ of a charged particle moving in a circular orbit in a magnetic field is given by the formula $r = \frac{mv}{qB}$.
Since kinetic energy $E = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mE}$.
Substituting this into the radius formula: $r = \frac{\sqrt{2mE}}{qB}$.
Given values: $m = 9 \times 10^{-31} \, kg$,$E = 7.2 \times 10^{-18} \, J$,$q = 1.6 \times 10^{-19} \, C$,and $B = 9 \times 10^{-5} \, Wb/m^2$.
$r = \frac{\sqrt{2 \times 9 \times 10^{-31} \times 7.2 \times 10^{-18}}}{1.6 \times 10^{-19} \times 9 \times 10^{-5}}$
$r = \frac{\sqrt{129.6 \times 10^{-49}}}{14.4 \times 10^{-24}} = \frac{\sqrt{12.96 \times 10^{-48}}}{14.4 \times 10^{-24}}$
$r = \frac{3.6 \times 10^{-24}}{14.4 \times 10^{-24}} = \frac{3.6}{14.4} = 0.25 \, m$.
Converting to centimeters: $0.25 \, m = 25 \, cm$.

(C) When a charged particle moves through a region containing both electric field $(E)$ and magnetic field $(B)$ without being deflected,the net Lorentz force acting on it must be zero.
The Lorentz force is given by $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
For the particle to pass undeflected,$\vec{F} = 0$,which implies $\vec{E} + \vec{v} \times \vec{B} = 0$,or $E = vB$ (assuming $\vec{v}$ is perpendicular to $\vec{B}$).
Therefore,the velocity of the electron is $v = \frac{E}{B}$.
Given $E = 20\,N/C$ and $B = 5\,T$,we have $v = \frac{20}{5} = 4\,m\,s^{-1}$.

(A) The force on a charged particle $q$ moving with velocity $\vec{v}$ in electric field $\vec{E}$ and magnetic field $\vec{B}$ is given by the Lorentz force equation: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
Since the particle is released from rest,its initial velocity $\vec{v} = 0$.
Initially,the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B}) = 0$.
The electric force $\vec{F}_e = q\vec{E}$ acts on the particle,causing it to accelerate in the direction of the electric field.
As the particle gains velocity $\vec{v}$ in the direction of $\vec{E}$,and since $\vec{E}$ and $\vec{B}$ are parallel,the velocity $\vec{v}$ remains parallel to the magnetic field $\vec{B}$ at all times.
Therefore,the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B}) = 0$ throughout the motion because the angle between $\vec{v}$ and $\vec{B}$ is $0^\circ$.
The particle continues to accelerate in a straight line due to the electric field.

(C) The magnetic force $\vec{F}_m$ acting on a charged particle moving in a magnetic field is given by $\vec{F}_m = Q(\vec{v} \times \vec{B})$.
Since the force $\vec{F}_m$ is always perpendicular to the velocity vector $\vec{v}$,the force is always directed towards the center of the circular path (centripetal force).
Work done $W$ is defined as the line integral of the force over the displacement: $W = \int \vec{F} \cdot d\vec{s}$.
Since the force $\vec{F}_m$ is always perpendicular to the displacement $d\vec{s}$ at every point on the circular path,the dot product $\vec{F}_m \cdot d\vec{s} = F_m ds \cos(90^\circ) = 0$.
Therefore,the total work done by the magnetic field on the particle in one full circle is $0$.

(B) The particle moves undeflected along the $x$-axis,which means the net Lorentz force acting on the particle is zero.
The Lorentz force is given by $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
For the particle to move undeflected,the electric force must be equal and opposite to the magnetic force: $q\vec{E} + q(\vec{v} \times \vec{B}) = 0$.
Given $\vec{v} = v\hat{i}$,$\vec{B} = B\hat{j}$,and $\vec{E} = -E\hat{k}$.
The magnetic force is $\vec{F}_m = q(v\hat{i} \times B\hat{j}) = qvB\hat{k}$.
The electric force is $\vec{F}_e = q(-E\hat{k}) = -qE\hat{k}$.
Equating the magnitudes: $qvB = qE$.
Therefore,$B = E/v$.
Substituting the given values: $B = \frac{10^4}{10} = 10^3 \, Wb/m^2$.

(D) When a charged particle enters a magnetic field perpendicularly,it follows a circular path.
The radius of the circular path is given by $r = \frac{mv}{qB}$.
Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mK}$.
Substituting this into the radius formula,we get $r = \frac{\sqrt{2mK}}{qB}$.
Since the electron and proton have the same kinetic energy $(K)$ and enter the same magnetic field $(B)$,the radius $r$ depends on $\frac{\sqrt{m}}{q}$.
Because the mass of a proton $(m_p)$ is much larger than the mass of an electron $(m_e)$,their radii will be different.
Therefore,both follow circular paths,but with different radii. Thus,options $A$,$B$,and $C$ are incorrect,making $D$ the correct statement.

(C) The radius $r$ of the circular path of a charged particle moving perpendicular to a magnetic field is given by the formula $r = \frac{mv}{qB}$.
Given that the specific charge (charge-to-mass ratio) is $\frac{q}{m} = 1.7 \times 10^{11} \text{ C/kg}$,we can rewrite the formula as $r = \frac{v}{(q/m)B}$.
Substituting the given values:
$v = 6 \times 10^7 \text{ m/s}$
$\frac{q}{m} = 1.7 \times 10^{11} \text{ C/kg}$
$B = 1.5 \times 10^{-2} \text{ T}$
$r = \frac{6 \times 10^7}{(1.7 \times 10^{11}) \times (1.5 \times 10^{-2})}$
$r = \frac{6 \times 10^7}{2.55 \times 10^9}$
$r \approx 2.35 \times 10^{-2} \text{ m}$
Converting to centimeters: $r = 2.35 \text{ cm}$.

(D) The magnetic force on a moving charge is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
Here,the electron has a negative charge $(q = -e)$.
The velocity $\vec{v}$ is in the $+X$ direction,and the magnetic field $\vec{B}$ is directed into the plane ($-Z$ direction).
Using the right-hand rule for the cross product $\vec{v} \times \vec{B}$,the direction is $(+X) \times (-Z) = +Y$.
Since the charge is negative,the force $\vec{F}$ is in the opposite direction,which is $-Y$.
Alternatively,using Fleming's left-hand rule: Point the index finger into the plane (magnetic field),the middle finger in the direction opposite to the electron's motion (conventional current direction,which is $-X$),then the thumb points in the direction of the force,which is $-Y$.

(C) The radius of a charged particle moving in a circular path in a uniform magnetic field is given by $r = \frac{\sqrt{2mK}}{qB}$,where $m$ is the mass,$K$ is the kinetic energy,and $q$ is the charge of the particle.
Since the radius $r$ and the magnetic field $B$ are the same for both particles,we have $q \propto \sqrt{mK}$,which implies $K \propto \frac{q^2}{m}$.
For a proton,$q_p = e$ and $m_p = m$. For an alpha particle,$q_{\alpha} = 2e$ and $m_{\alpha} = 4m$.
Taking the ratio of kinetic energies: $\frac{K_{\alpha}}{K_p} = \left( \frac{q_{\alpha}}{q_p} \right)^2 \times \frac{m_p}{m_{\alpha}}$.
Substituting the values: $\frac{K_{\alpha}}{8} = \left( \frac{2e}{e} \right)^2 \times \frac{m}{4m} = 4 \times \frac{1}{4} = 1$.
Therefore,$K_{\alpha} = 8\, eV$.

(C) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$.
Since the speed $v$ and magnetic field $B$ are constant for all particles,we have $r \propto \frac{m}{q}$,or $r \propto \frac{1}{(q/m)}$.
For the given particles:
$1$. Electron $(e^-)$: charge $q$,mass $m_e$.
$2$. Proton $(p^+)$: charge $q$,mass $m_p \approx 1836 m_e$.
$3$. Deuteron $(d)$: charge $q$,mass $m_d \approx 2 m_p$.
$4$. Alpha particle $(\alpha)$: charge $2q$,mass $m_\alpha \approx 4 m_p$.
Comparing the charge-to-mass ratios $(q/m)$:
$(q/m)_d = q / (2 m_p) = 0.5 (q/m_p)$
$(q/m)_\alpha = 2q / (4 m_p) = 0.5 (q/m_p)$
Since $(q/m)_d = (q/m)_\alpha$,it follows that $R_d = R_\alpha$.

(D) The magnetic Lorentz force on a moving charge is given by the formula $\vec{F} = q(\vec{v} \times \vec{B})$.
Here,the charge of the electron is $q = -e$ (negative).
The velocity vector is $\vec{v} = v \hat{i}$ and the magnetic field vector is $\vec{B} = B \hat{j}$.
Substituting these into the formula,we get $\vec{F} = -e(v \hat{i} \times B \hat{j})$.
Since $\hat{i} \times \hat{j} = \hat{k}$,the force becomes $\vec{F} = -evB \hat{k}$.
Therefore,the force is directed along the negative $z$-direction.

(D) The magnetic field $\overrightarrow{B}$ produced by a circular current-carrying coil is directed along its axis at all points on the axis.
Since the electron is projected along the axis,its velocity vector $\overrightarrow{v}$ is parallel to the magnetic field vector $\overrightarrow{B}$.
The magnetic force on a moving charge is given by $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
Since $\overrightarrow{v}$ and $\overrightarrow{B}$ are parallel,the angle $\theta$ between them is $0^o$ or $180^o$.
Therefore,$\overrightarrow{F} = qvB \sin(\theta) = qvB \sin(0^o) = 0$.
Thus,the electron experiences no force.

(D) The magnetic Lorentz force acting on a moving charge $q$ in a magnetic field $B$ is given by the formula $F_m = q(v \times B)$,which has a magnitude of $F_m = qvB \sin \theta$.
Since the charge is stationary,its velocity $v = 0$.
Substituting $v = 0$ into the formula,we get $F_m = q(0)B \sin \theta = 0$.
Therefore,a stationary charge does not experience any magnetic force,regardless of the strength of the magnetic field.

(D) The time period $T$ of a charged particle moving in a circular path in a uniform magnetic field is given by the formula: $T = \frac{2\pi m}{qB}$.
Given:
Mass of electron $m = 9.1 \times 10^{-31}\, kg$
Charge $q = 1.6 \times 10^{-19}\, C$
Magnetic field $B = 3.534 \times 10^{-5}\, T$
Substituting the values:
$T = \frac{2 \times 3.14 \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 3.534 \times 10^{-5}}$
$T = \frac{57.148 \times 10^{-31}}{5.6544 \times 10^{-24}}$
$T \approx 10.106 \times 10^{-7}\, s \approx 1 \times 10^{-6}\, s = 1\,\mu s$.
Therefore,the correct option is $D$.

(B) When a charged particle moves perpendicular to a magnetic field,the magnetic Lorentz force provides the necessary centripetal force for circular motion.
Equating the magnetic force to the centripetal force:
$qvB = \frac{mv^2}{r}$
From this,we can find the velocity $v$ or the radius $r$:
$v = \frac{qBr}{m}$
The time period $T$ for one complete revolution is the circumference divided by the speed:
$T = \frac{2\pi r}{v}$
Substituting the expression for $v$:
$T = \frac{2\pi r}{(qBr/m)} = \frac{2\pi m}{qB}$
Thus,the time taken to complete one revolution is $\frac{2\pi m}{qB}$.