Motion of Charged Particle In Magnetic Field Questions in English

(D) The $SI$ unit of magnetic field induction $(B)$ is Tesla $(T)$.
One Tesla is defined as the magnetic field that exerts a force of $1 \ N$ on a charge of $1 \ C$ moving with a velocity of $1 \ m/s$ perpendicular to the field.

(B) The unit of magnetic field induction $(B)$ in the $CGS$ system is Gauss $(G)$.
$1 \text{ Gauss} = 10^{-4} \text{ Tesla}$.
Therefore,Gauss is the unit of magnetic field $(B)$.

(D) The magnetic force on a moving charge is given by $\vec{F}_m = q(\vec{v} \times \vec{B})$. Since the electron is moving parallel to the magnetic field,the angle between $\vec{v}$ and $\vec{B}$ is $0^\circ$,so $\vec{F}_m = 0$.
The electric force on the electron is given by $\vec{F}_e = q\vec{E}$. Since the charge of an electron is negative $(q = -e)$,the electric force acts in the direction opposite to the electric field $\vec{E}$.
As the electron is projected in the direction of the electric field,the electric force acts in the opposite direction to its velocity. This force acts as a retarding force,causing the electron to decelerate. Therefore,the magnitude of the electron's velocity will decrease.

(C) The radius $r$ of a charged particle moving in a magnetic field $B$ after being accelerated through a potential difference $V$ is given by the formula:
$r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$
Since the charges $q$,potential difference $V$,and magnetic field $B$ are the same for both particles,we have $r \propto \sqrt{m}$.
Therefore,$\frac{R_1}{R_2} = \sqrt{\frac{m_X}{m_Y}}$.
Squaring both sides,we get $\frac{m_X}{m_Y} = (\frac{R_1}{R_2})^2$.

(A) When a charged particle enters a uniform magnetic field perpendicularly,it follows a circular path.
The radius $r$ of the circular path is given by the formula $r = \frac{mv}{Bq}$.
Here,$v = 2 \times 10^5 \ m/s$,$B = 4 \times 10^{-2} \ T$,and the specific charge $\frac{q}{m} = 5 \times 10^7 \ C/kg$.
We can rewrite the formula as $r = \frac{v}{(\frac{q}{m})B}$.
Substituting the given values:
$r = \frac{2 \times 10^5}{(5 \times 10^7) \times (4 \times 10^{-2})}$
$r = \frac{2 \times 10^5}{20 \times 10^5}$
$r = \frac{2}{20} = 0.1 \ m$.

(B) When a charged particle with charge $q$ and momentum $p$ moves in a uniform magnetic field $B$ perpendicular to the field,it follows a circular path.
The magnetic Lorentz force provides the necessary centripetal force:
$F_m = F_c$
$qvB = \frac{mv^2}{r}$
Since momentum $p = mv$,we can write $qvB = \frac{pv}{r}$.
Solving for the radius $r$,we get:
$r = \frac{p}{qB}$
From this expression,it is clear that the radius $r$ is directly proportional to the momentum $p$ of the particle $(r \propto p)$.

(B) When a charged particle moves in a magnetic field perpendicular to its velocity,it follows a circular path.
The radius $r$ of the circular path is given by the formula: $r = \frac{mv}{qB}$.
Rearranging the formula to solve for the magnetic field intensity $B$,we get: $B = \frac{mv}{qr}$.
Given values:
Mass of electron $(m)$ = $9 \times 10^{-31} \ kg$
Velocity $(v)$ = $10^6 \ m/s$
Charge $(q)$ = $1.6 \times 10^{-19} \ C$
Radius $(r)$ = $0.10 \ m$
Substituting these values into the formula:
$B = \frac{9 \times 10^{-31} \times 10^6}{1.6 \times 10^{-19} \times 0.1}$
$B = \frac{9 \times 10^{-25}}{0.16 \times 10^{-19}}$
$B = \frac{9}{0.16} \times 10^{-6} \ T$
$B = 56.25 \times 10^{-6} \ T = 5.625 \times 10^{-5} \ T$.
Rounding to the nearest option,we get $B = 5.6 \times 10^{-5} \ T$.

(C) The radius $r$ of a charged particle moving in a uniform magnetic field $B$ with kinetic energy $K$ is given by the formula $r = \frac{\sqrt{2mK}}{qB}$.
Since $K$ and $B$ are the same for both particles,we have $r \propto \frac{\sqrt{m}}{q}$.
For the proton: $m_p = m$ and $q_p = e$. Thus,$r_p \propto \frac{\sqrt{m}}{e}$.
For the $\alpha$-particle: $m_{\alpha} = 4m$ and $q_{\alpha} = 2e$. Thus,$r_{\alpha} \propto \frac{\sqrt{4m}}{2e} = \frac{2\sqrt{m}}{2e} = \frac{\sqrt{m}}{e}$.
Comparing the two,we find that $r_p = r_{\alpha}$.
Therefore,both particles will move in a circular path with the same radius.

(C) The magnetic Lorentz force acting on a charged particle moving with velocity $\overrightarrow{v}$ in a magnetic field $\overrightarrow{B}$ is given by the formula: $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
According to the properties of the cross product,the resulting vector $\overrightarrow{F}$ is always perpendicular to the plane containing the vectors $\overrightarrow{v}$ and $\overrightarrow{B}$.
Therefore,the force is perpendicular to both the velocity of the particle and the magnetic field.

(C) When a charged particle with charge $q$ and mass $m$ moves with velocity $v$ in a uniform magnetic field $B$,the magnetic Lorentz force is given by $F = q(v \times B)$.
If the velocity $v$ is perpendicular to the magnetic field $B$,the angle between them is $90^\circ$.
The magnitude of the force is $F = qvB \sin(90^\circ) = qvB$.
This force acts perpendicular to both the velocity and the magnetic field,acting as a centripetal force.
Since the force is always perpendicular to the velocity,the speed of the particle remains constant,but the direction changes continuously.
This constant perpendicular force results in a uniform circular motion.
Therefore,the path of the charged particle is a circle.

(D) When a charged particle enters a magnetic field with a velocity $v$ at an angle $\theta$ (where $\theta \neq 0^\circ, 90^\circ, 180^\circ$) with respect to the magnetic field $B$,the velocity can be resolved into two components:
$1$. The component $v \sin \theta$ perpendicular to the magnetic field,which provides the necessary centripetal force for circular motion.
$2$. The component $v \cos \theta$ parallel to the magnetic field,which causes the particle to move linearly along the field lines.
The combination of these two motions results in a helical path.

(C) The force on a moving charged particle in a magnetic field is given by the Lorentz force formula: $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
Here,the velocity $\overrightarrow{v}$ of the positively charged particles is directed downwards towards the Earth.
The Earth's magnetic field $\overrightarrow{B}$ at the equator is directed from South to North.
Using the right-hand rule for the cross product $\overrightarrow{v} \times \overrightarrow{B}$ (where $\overrightarrow{v}$ is downward and $\overrightarrow{B}$ is towards the North),the direction of the force $\overrightarrow{F}$ is towards the East.
Therefore,the positively charged particles are deflected towards the East.

(D) The force on a charged particle moving in a magnetic field is given by $F = qvB \sin \theta$. Since the proton moves perpendicular to the field,$\theta = 90^\circ$,so $F = qvB$.
Given: Energy $E = 2 \, MeV = 2 \times 10^6 \times 1.6 \times 10^{-19} \, J = 3.2 \times 10^{-13} \, J$.
Mass of proton $m \approx 1.67 \times 10^{-27} \, kg$.
Velocity $v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 3.2 \times 10^{-13}}{1.67 \times 10^{-27}}} \approx \sqrt{3.83 \times 10^{14}} \approx 1.96 \times 10^7 \, m/s$.
Force $F = (1.6 \times 10^{-19} \, C) \times (1.96 \times 10^7 \, m/s) \times (2.5 \, T) \approx 7.84 \times 10^{-12} \, N$.
Rounding to the nearest provided option,the force is $7.6 \times 10^{-12} \, N$.

(B) The magnetic force $\overrightarrow{F}$ on a charged particle moving with velocity $\overrightarrow{v}$ in a magnetic field $\overrightarrow{B}$ is given by the Lorentz force formula: $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
Since the cross product $\overrightarrow{v} \times \overrightarrow{B} = vB \sin(\theta) \hat{n}$,where $\theta$ is the angle between $\overrightarrow{v}$ and $\overrightarrow{B}$,the force is zero when $\sin(\theta) = 0$.
This occurs when $\theta = 0^\circ$ or $\theta = 180^\circ$,which means $\overrightarrow{v}$ and $\overrightarrow{B}$ are parallel or anti-parallel.
Therefore,the force is zero if $\overrightarrow{B}$ and $\overrightarrow{v}$ are parallel.

(B) The force experienced by a charged particle moving in a magnetic field is given by the Lorentz force formula: $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
Here,the velocity of the proton is along the $Z$-axis,so $\overrightarrow{v} = v\hat{k}$.
The magnetic field is along the $X$-axis,so $\overrightarrow{B} = B\hat{i}$.
Substituting these into the formula: $\overrightarrow{F} = q(v\hat{k} \times B\hat{i})$.
Using the cross product rules for unit vectors $(\hat{k} \times \hat{i} = \hat{j})$,we get $\overrightarrow{F} = qvB\hat{j}$.
Since $\hat{j}$ represents the $Y$-axis,the force acts along the $Y$-axis.

(A) The radius of a charged particle moving in a magnetic field is given by $r = \frac{\sqrt{2mK}}{qB}$,where $K$ is the kinetic energy.
For the same radius $r$ and same magnetic field $B$,we have $r \propto \frac{\sqrt{mK}}{q}$.
Squaring both sides,$r^2 \propto \frac{mK}{q^2}$,which implies $K \propto \frac{q^2}{m}$.
Therefore,$\frac{K_p}{K_\alpha} = \left( \frac{q_p}{q_\alpha} \right)^2 \times \frac{m_\alpha}{m_p}$.
Given $q_p = e$,$q_\alpha = 2e$,$m_p = m$,and $m_\alpha = 4m$.
Substituting these values: $\frac{1\, MeV}{K_\alpha} = \left( \frac{e}{2e} \right)^2 \times \frac{4m}{m} = \left( \frac{1}{4} \right) \times 4 = 1$.
Thus,$K_\alpha = 1\, MeV$.

(C) The radius $r$ of a charged particle moving in a uniform magnetic field $B$ is given by the formula $r = \frac{mv}{qB}$,where $m$ is the mass,$v$ is the velocity,and $q$ is the charge of the particle.
From this formula,we can see that $r \propto \frac{1}{B}$.
Let the initial radius be $r_1 = r$ at magnetic field $B_1 = B$.
Let the final radius be $r_2$ at magnetic field $B_2 = B/2$.
Using the proportionality $r_1 B_1 = r_2 B_2$,we get:
$r \times B = r_2 \times (B/2)$
$r_2 = \frac{r \times B}{B/2} = 2r$.
Therefore,the radius increases to $2r$.

(C) The time period of a charged particle moving in a uniform magnetic field is given by $T = \frac{2\pi m}{qB}$.
For a proton,the time taken for $5$ revolutions is $25 \ \mu s$,so the time period $T_p = \frac{25 \ \mu s}{5} = 5 \ \mu s$.
For an $\alpha$-particle,the mass is $m_{\alpha} = 4m_p$ and the charge is $q_{\alpha} = 2q_p$.
Using the ratio: $\frac{T_{\alpha}}{T_p} = \frac{m_{\alpha}}{m_p} \times \frac{q_p}{q_{\alpha}} = \frac{4m_p}{m_p} \times \frac{q_p}{2q_p} = 4 \times \frac{1}{2} = 2$.
Therefore,$T_{\alpha} = 2 \times T_p = 2 \times 5 \ \mu s = 10 \ \mu s$.

(A) The magnetic force acting on a charged particle moving in a magnetic field is given by $F = qvB \sin(\theta)$.
Since the proton enters perpendicular to the magnetic field,$\theta = 90^\circ$,so $\sin(90^\circ) = 1$.
Thus,the force is $F = qvB$.
According to Newton's second law,$F = ma$,therefore $ma = qvB$.
The acceleration $a$ is given by $a = \frac{qvB}{m}$.
Substituting the given values:
$a = \frac{(1.6 \times 10^{-19} \, C) \times (3.4 \times 10^7 \, m/s) \times (2 \, Wb/m^2)}{1.67 \times 10^{-27} \, kg}$.
$a = \frac{10.88 \times 10^{-12}}{1.67 \times 10^{-27}} \, m/s^2$.
$a \approx 6.515 \times 10^{15} \, m/s^2$.
Rounding to the given options,$a = 6.5 \times 10^{15} \, m/s^2$.

(C) The time period $T$ of a charged particle moving in a circular path in a magnetic field is given by the formula $T = \frac{2\pi r}{v}$.
Given:
Radius $r = 0.45\,m$
Speed $v = 2.6 \times 10^7\,m/s$
Substituting the values into the formula:
$T = \frac{2 \times 3.14 \times 0.45}{2.6 \times 10^7}$
$T = \frac{2.826}{2.6 \times 10^7}$
$T \approx 1.0869 \times 10^{-7}\,s$
Rounding to two significant figures,we get $T \approx 1.1 \times 10^{-7}\,s$.

(B) The force on a charged particle moving in a magnetic field is given by $F = qvB \sin \theta$. Since the particle moves vertically downwards and the field is horizontal (south to north),the angle $\theta = 90^\circ$,so $\sin 90^\circ = 1$. Thus,$F = qvB$.
Given kinetic energy $K = 5 \ MeV = 5 \times 10^6 \times 1.6 \times 10^{-19} \ J = 8 \times 10^{-13} \ J$.
Using $K = \frac{1}{2}mv^2$,we find velocity $v = \sqrt{\frac{2K}{m}}$.
Substituting $v$ into the force equation: $F = qB \sqrt{\frac{2K}{m}}$.
$F = (1.6 \times 10^{-19}) \times 1.5 \times \sqrt{\frac{2 \times 8 \times 10^{-13}}{1.7 \times 10^{-27}}}$.
$F = 2.4 \times 10^{-19} \times \sqrt{9.41 \times 10^{14}} \approx 2.4 \times 10^{-19} \times 3.067 \times 10^7$.
$F \approx 7.36 \times 10^{-12} \ N$,which is approximately $7.4 \times 10^{-12} \ N$.

(C) 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: $F = q(v \times B)$.
Since the electron is stationary,its velocity $v = 0$.
Therefore,the magnetic force $F = q(0 \times B) = 0$.
Because no net force acts on the stationary electron,it remains stationary.

(C) The radius $r$ of a circular path of a charged particle moving in a uniform magnetic field is given by the formula $r = \frac{mv}{qB}$,where $m$ is the mass,$v$ is the speed,$q$ is the charge,and $B$ is the magnetic field strength.
Since $m$,$q$,and $B$ are constant,the radius is directly proportional to the speed,i.e.,$r \propto v$.
Given the initial radius $r_1 = 2\, cm$ and the initial speed $v_1 = v$.
If the speed is doubled,$v_2 = 2v$.
Therefore,the new radius $r_2$ is given by $\frac{r_2}{r_1} = \frac{v_2}{v_1} = \frac{2v}{v} = 2$.
Thus,$r_2 = 2 \times r_1 = 2 \times 2\, cm = 4\, cm$.

(D) The radius of a circular path of a charged particle in a 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.
From this,we can write $K = \frac{r^2 q^2 B^2}{2m}$.
Since $r$ and $B$ are the same for both particles,$K \propto \frac{q^2}{m}$.
For a deuteron $(d)$,$q_d = e$ and $m_d = 2m_p$. For a proton $(p)$,$q_p = e$ and $m_p = m_p$.
Taking the ratio: $\frac{K_p}{K_d} = \left( \frac{q_p}{q_d} \right)^2 \times \left( \frac{m_d}{m_p} \right) = \left( \frac{e}{e} \right)^2 \times \left( \frac{2m_p}{m_p} \right) = 1^2 \times 2 = 2$.
Therefore,$K_p = 2 \times K_d = 2 \times 50 \, keV = 100 \, keV$.

(B) The magnetic force on a charged particle is given by $F = q(v \times B) = qvB \sin \theta$.
For the proton,the velocity is perpendicular to the magnetic field,so $\theta = 90^\circ$. The force $F = qvB \sin 90^\circ = qvB$ acts as a centripetal force,causing the proton to move in a circular path with constant speed.
For the electron,the velocity is parallel to the magnetic field lines,so $\theta = 0^\circ$. The force $F = qvB \sin 0^\circ = 0$. Since no force acts on the electron,its motion remains unaffected (it continues to move in a straight line).

(D) 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 (let this be the $+x$ direction).
The magnetic field $\vec{B}$ is vertically downward (let this be the $-z$ direction).
Using the cross product: $\vec{v} \times \vec{B} = (v \hat{i}) \times (-B \hat{k}) = -vB(\hat{i} \times \hat{k}) = -vB(-\hat{j}) = vB \hat{j}$ (which is North).
Since the electron has a negative charge,the force $\vec{F} = -e(vB \hat{j}) = -evB \hat{j}$.
The direction $-\hat{j}$ corresponds to the South direction.
Therefore,the force on the electron is towards the South.

(C) The magnetic force on a charged particle moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
This force is $\vec{F} = qvB \sin \theta$,where $\theta$ is the angle between the velocity vector and the magnetic field vector.
If the particle moves across the magnetic field lines,$\theta \neq 0$ and $\theta \neq 180^{\circ}$,so $\sin \theta \neq 0$,resulting in a non-zero magnetic force.
If the particle moves along the magnetic field lines,$\theta = 0^{\circ}$ or $180^{\circ}$,so $\sin \theta = 0$,resulting in zero magnetic force.
Therefore,the correct option is $C$.

(D) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Given: $q = 10^{-12} \, C$,$\vec{v} = 10^5 \hat{i} \, m/s$,and $\vec{F} = 10^{-10} \hat{j} \, N$.
The magnitude of the force is $F = qvB \sin \theta$.
For the minimum magnetic field,we assume $\sin \theta = 1$ (i.e.,$\theta = 90^\circ$),so $F = qvB$.
Thus,$B_{\min} = \frac{F}{qv} = \frac{10^{-10}}{10^{-12} \times 10^5} = \frac{10^{-10}}{10^{-7}} = 10^{-3} \, T$.
Using the right-hand rule,since $\vec{v}$ is in $\hat{i}$ and $\vec{F}$ is in $\hat{j}$,the magnetic field $\vec{B}$ must be in the $\hat{k}$ direction (since $\hat{i} \times \hat{k} = -\hat{j}$,we check the direction: $\vec{F} = q(\vec{v} \times \vec{B}) \implies \hat{j} = \hat{i} \times \hat{B} \implies \hat{B} = \hat{k}$).

(D) When a charged particle of charge $q$ is accelerated from rest through a potential difference $V$, its kinetic energy $K$ is given by the work-energy theorem as $K = qV$.
Since the potential difference $V$ is the same for all particles, the kinetic energy is directly proportional to the charge of the particle, i.e., $K \propto q$.
For a proton $(p)$, deuteron $(d)$, and $\alpha$-particle $(\alpha)$, the charges are:
$q_p = e$
$q_d = e$
$q_\alpha = 2e$
Therefore, the ratio of their kinetic energies is:
$K_p : K_d : K_\alpha = q_p : q_d : q_\alpha = e : e : 2e = 1 : 1 : 2$.

(D) Option $(D)$ is correct.
According to the Lorentz force law,the magnetic force acting on a charged particle is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Since the magnetic force $\vec{F}$ is always perpendicular to the velocity vector $\vec{v}$,the work done by the magnetic force on the particle 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 by the net force.
Since the work done by the magnetic force is zero,the kinetic energy of the charged particle remains constant,even though a magnetic force acts on it.

(B) The radius of a circular path for 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 $mv = \sqrt{2mK}$.
Substituting this into the radius formula,we get $r = \frac{\sqrt{2mK}}{qB}$.
Given that kinetic energy $K$ and magnetic field $B$ are the same for both particles,we have $r \propto \frac{\sqrt{m}}{q}$.
Since the magnitude of charge for an electron and a proton is the same $(|q_e| = |q_p| = e)$,the ratio of the radii is $\frac{r_e}{r_p} = \sqrt{\frac{m_e}{m_p}}$.
Because the mass of an electron $m_e$ is much smaller than the mass of a proton $m_p$ $(m_e < m_p)$,it follows that $r_e < r_p$.

(C) The magnetic field $\vec{B}$ is along the $Y$-axis. The velocity $\vec{v}$ makes an angle of $60^\circ$ with the $X$-axis,so it makes an angle $\theta = 90^\circ - 60^\circ = 30^\circ$ with the $Y$-axis (the direction of $\vec{B}$).
Since the velocity has a component parallel to the magnetic field,the path of the proton is a helix.
The radius of the helix is given by $r = \frac{mv \sin \theta}{qB}$,where $\theta = 30^\circ$ is the angle between $\vec{v}$ and $\vec{B}$.
$r = \frac{1.67 \times 10^{-27} \times 2 \times 10^6 \times \sin 30^\circ}{1.6 \times 10^{-19} \times 0.104} = \frac{1.67 \times 10^{-21} \times 0.5}{1.664 \times 10^{-20}} \approx 0.1 \, m$.
The time period is $T = \frac{2\pi m}{qB} = \frac{2\pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.104} \approx 2\pi \times 10^{-7} \, s$.

(A) The magnetic force provides the necessary centripetal force for circular motion: $\frac{mv^2}{R} = qvB$,which simplifies to $R = \frac{mv}{qB}$.
Since the kinetic energy $E = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mE}$.
Substituting this into the radius formula: $R = \frac{\sqrt{2mE}}{qB}$.
For a proton,${R_p} = \frac{\sqrt{2{m_p}E}}{{q_p}B}$. For a deuteron,${R_d} = \frac{\sqrt{2{m_d}E}}{{q_d}B}$.
Since a deuteron has the same charge as a proton $(q_d = q_p)$ and its mass is twice that of a proton $(m_d = 2{m_p})$,we get:
$\frac{{R_d}}{{R_p}} = \sqrt{\frac{m_d}{m_p}} = \sqrt{\frac{2{m_p}}{{m_p}}} = \sqrt{2}$.
Therefore,${R_d} = \sqrt{2} {R_p}$.

(C) The total force acting on a charged particle moving in both electric and magnetic fields is given by the Lorentz force equation: $\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B})$.
For the proton to move undeflected,the net force must be zero,meaning $\overrightarrow{F} = 0$.
This implies $q\overrightarrow{E} = -q(\overrightarrow{v} \times \overrightarrow{B})$,or $\overrightarrow{E} = -(\overrightarrow{v} \times \overrightarrow{B})$.
This condition is satisfied when $\overrightarrow{E}$,$\overrightarrow{B}$,and $\overrightarrow{v}$ are mutually perpendicular to each other.
In this configuration,the magnitude of the electric force $|F_e| = qE$ must equal the magnitude of the magnetic force $|F_m| = qvB$.
Therefore,$qE = qvB$,which simplifies to $v = \frac{E}{B}$.

(B) The time period $T$ of a charged particle moving in a uniform magnetic field $B$ is given by the formula $T = \frac{2\pi m}{qB}$.
Since the charge magnitude $q$ (for both proton and electron) and the magnetic field $B$ are the same,the time period is directly proportional to the mass of the particle,i.e.,$T \propto m$.
We know that the mass of a proton $(m_p)$ is significantly greater than the mass of an electron $(m_e)$,i.e.,$m_p > m_e$.
Therefore,the time period of the proton will be greater than the time period of the electron.

(A) When a charged particle moves in a transverse magnetic field,the magnetic Lorentz force provides the necessary centripetal force for circular motion.
$F_m = F_c$
$evB = \frac{mv^2}{r}$
Rearranging the terms to solve for the specific charge $e/m$:
$eB = \frac{mv}{r}$
$\frac{e}{m} = \frac{v}{Br}$
Therefore,the correct option is $A$.

(A) When a charged particle moves through mutually perpendicular electric and magnetic fields without being deviated,the net Lorentz force acting on it is zero.
This implies that the electric force is equal in magnitude and opposite in direction to the magnetic force.
Mathematically,$F_e = F_m$.
Since $F_e = qE$ and $F_m = qvB$,we have $qE = qvB$.
Therefore,$E = vB$.
Given $v = 5 \times 10^6 \, m/s$ and $B = 0.02 \, T$.
Substituting these values,$E = (5 \times 10^6) \times 0.02 = 10^5 \, V/m$.

(D) The magnetic force $F$ acting on a charged particle moving in a magnetic field is given by $F = qvB \sin \theta$. Since the electron moves in a circular orbit,$\theta = 90^\circ$,so $F = evB$.
$F = (1.6 \times 10^{-19} \, C) \times (4 \times 10^6 \, m/s) \times (2 \times 10^{-1} \, T) = 1.28 \times 10^{-13} \, N$.
The centripetal force required for circular motion is provided by the magnetic force: $\frac{mv^2}{r} = evB$.
Rearranging for the radius $r$,we get $r = \frac{mv}{eB}$.
$r = \frac{(9 \times 10^{-31} \, kg) \times (4 \times 10^6 \, m/s)}{(1.6 \times 10^{-19} \, C) \times (2 \times 10^{-1} \, T)} = \frac{36 \times 10^{-25}}{3.2 \times 10^{-20}} = 1.125 \times 10^{-4} \, m \approx 1.1 \times 10^{-4} \, m$.

(C) When a charged particle enters a magnetic field with velocity $\vec{v}$,it experiences a magnetic Lorentz force given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Since the force $\vec{F}$ is always perpendicular to the velocity $\vec{v}$,the work done by the magnetic force on the electron is $W = \int \vec{F} \cdot d\vec{s} = \int \vec{F} \cdot \vec{v} dt = 0$.
According to the work-energy theorem,the change in kinetic energy is equal to the work done. Since the work done is $0$,the kinetic energy remains constant.
Therefore,the speed of the electron remains the same,although its direction of motion changes.

(B) The force on a moving charge is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
For an electron,the charge $q$ is negative $(-e)$.
Given: Velocity $\vec{v}$ is in the North direction,and Force $\vec{F}$ is in the vertically upward direction.
Using the right-hand rule for the cross product $\vec{v} \times \vec{B}$,if $\vec{v}$ is North and $\vec{F}$ (which is $-e(\vec{v} \times \vec{B})$) is Upward,then $(\vec{v} \times \vec{B})$ must be in the downward direction.
If we point our fingers in the North direction and curl them such that the thumb points downward,the fingers point towards the West.
Therefore,the magnetic field $\vec{B}$ is in the West direction.

(D) The magnetic field inside a long current-carrying solenoid is directed along its axis.
Since the proton is moving along the axis of the solenoid,its velocity vector $v$ is parallel to the magnetic field vector $B$.
The magnetic force $F$ on a charged particle is given by the Lorentz force formula: $F = q(v \times B)$.
Since $v$ and $B$ are parallel,the angle $\theta$ between them is $0^\circ$ or $180^\circ$.
Therefore,the magnetic force $F = qvB \sin(\theta) = qvB \sin(0^\circ) = 0$.
As there is no magnetic force acting on the proton,it will continue to move along the same path with no change in its velocity.

(A) When a charged particle moves in a uniform magnetic field $B$ perpendicular to its velocity,it experiences a magnetic Lorentz force which provides the necessary centripetal force for circular motion.
$F_m = F_c$
$qvB = \frac{mv^2}{r}$
$v = \frac{qBr}{m}$
The time period $T$ of revolution is the distance covered in one circle divided by the speed:
$T = \frac{2\pi r}{v} = \frac{2\pi r}{(qBr/m)} = \frac{2\pi m}{qB}$
The frequency of revolution $\nu$ is the reciprocal of the time period:
$\nu = \frac{1}{T} = \frac{qB}{2\pi m}$

(B) The kinetic energy $K$ gained by an electron accelerated through a potential difference $V$ is given by $K = qV$.
The radius $r$ of the circular path of a charged particle in a uniform magnetic field $B$ is given by $r = \frac{mv}{qB}$.
Since $K = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2qV}{m}}$.
Substituting $v$ into the radius formula: $r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.
Given values: $V = 12000\, V$,$B = 10^{-3}\, T$,$m = 9 \times 10^{-31}\, kg$,$q = 1.6 \times 10^{-19}\, C$.
$r = \frac{1}{10^{-3}} \sqrt{\frac{2 \times 9 \times 10^{-31} \times 12000}{1.6 \times 10^{-19}}} = 10^3 \sqrt{\frac{216 \times 10^{-28}}{1.6 \times 10^{-19}}} = 10^3 \sqrt{135 \times 10^{-9}} = 10^3 \sqrt{1.35 \times 10^{-7}} \approx 0.367\, m$.
Therefore,$r = 36.7\, cm$.

(C) The radius of a circular path of a charged particle accelerated through a potential difference $V$ in a magnetic field $B$ is given by $r = \frac{1}{B}\sqrt{\frac{2mV}{q}}$.
Since $B$ and $V$ are the same for both particles,we have $r \propto \sqrt{\frac{m}{q}}$.
Therefore,$\frac{R_1}{R_2} = \sqrt{\frac{m_X}{m_Y} \cdot \frac{q_Y}{q_X}}$.
Given that $q_Y = 2q_X$,we substitute this into the equation:
$\frac{R_1}{R_2} = \sqrt{\frac{m_X}{m_Y} \cdot \frac{2q_X}{q_X}} = \sqrt{\frac{m_X}{m_Y} \cdot 2}$.
Squaring both sides,we get $\frac{R_1^2}{R_2^2} = \frac{m_X}{m_Y} \cdot 2$.
Thus,the ratio of the mass of $X$ to that of $Y$ is $\frac{m_X}{m_Y} = \frac{R_1^2}{2R_2^2}$.

(D) The magnetic force on a moving charge is given by the Lorentz force formula: $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
Given: $q = 10^{-11} \, C$,$\overrightarrow{v} = 10^8 \hat{j} \, m/s$,and $\overrightarrow{B} = 0.5 \hat{i} \, T$.
Substituting these values into the formula:
$\overrightarrow{F} = 10^{-11} \times (10^8 \hat{j} \times 0.5 \hat{i})$
$\overrightarrow{F} = 10^{-11} \times 10^8 \times 0.5 \times (\hat{j} \times \hat{i})$
Since $\hat{j} \times \hat{i} = -\hat{k}$,we get:
$\overrightarrow{F} = 0.5 \times 10^{-3} \times (-\hat{k})$
$\overrightarrow{F} = 5 \times 10^{-4} \, N$ along $-\hat{k}$.

(B) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path.
The magnetic force provides the necessary centripetal force: $qvB = \frac{mv^2}{r}$.
Solving for the radius $r$,we get $r = \frac{mv}{qB}$.
As shown in the figure,the particle enters at $x=0$ and moves in a semi-circular path before exiting the region $x > 0$. The maximum distance $d$ it penetrates into the region $x > 0$ is equal to the radius of this circular path.
Therefore,$d = r = \frac{mv}{qB}$.

(C) The magnetic Lorentz force acting on a charged particle moving with velocity $v$ in a magnetic field $B$ is given by the vector product: $F = q(v \times B)$.
The magnitude of this force is given by $F = qvB \sin \theta$,where $\theta$ is the angle between the velocity vector $v$ and the magnetic field vector $B$.
For the force $F$ to be maximum,the value of $\sin \theta$ must be maximum.
The maximum value of $\sin \theta$ is $1$,which occurs when $\theta = 90^{\circ}$.
Therefore,the force is maximum when the velocity $v$ and the magnetic field $B$ are perpendicular to each other.

(D) When a charged particle enters a magnetic field $B$ at an angle $\theta$ (where $\theta \neq 0^\circ, 90^\circ, 180^\circ$),its velocity $v$ can be resolved into two components:
$1$. The component $v \cos \theta$ parallel to the magnetic field,which causes the particle to move in a straight line along the direction of the field.
$2$. The component $v \sin \theta$ perpendicular to the magnetic field,which causes the particle to move in a circular path.
Since the particle has both a parallel and a perpendicular velocity component,the resultant path is a helix. Given $\theta = 45^\circ$,the path is a helix.

(B) The magnetic force acting on a charged particle moving perpendicularly to a magnetic field provides the centripetal force: $qvB = \frac{mv^2}{r}$.
Rearranging for the radius of the path,we get $r = \frac{mv}{qB}$.
Since kinetic energy $K = \frac{1}{2}mv^2$,we can write $mv = \sqrt{2mK}$.
Substituting this into the radius formula,we get $r = \frac{\sqrt{2mK}}{qB}$.
For the same kinetic energy $K$ and same charge magnitude $q$,the radius is proportional to the square root of the mass: $r \propto \sqrt{m}$.
Since the mass of a proton $(m_p)$ is much greater than the mass of an electron $(m_e)$,the radius of the proton's path $(r_p)$ will be greater than the radius of the electron's path $(r_e)$.
$A$ larger radius of curvature implies that the path is less curved. Therefore,the trajectory of the proton is less curved.