Motion of Charged Particle In Magnetic Field Questions in English

$(A)$ Magnetic flux density,$B = 1.5 \,Wb \,m^{-2}$
Velocity of proton,$v = 2 \times 10^7 \,ms^{-1}$
Angle,$\theta = 30^{\circ}$
Charge on proton,$q = 1.6 \times 10^{-19} \,C$
The magnetic force $F$ on a moving charge is given by $F = Bqv \sin \theta$.
Substituting the values:
$F = (1.5) \times (1.6 \times 10^{-19}) \times (2 \times 10^7) \times \sin 30^{\circ}$
$F = 1.5 \times 1.6 \times 2 \times 10^{-12} \times 0.5$
$F = 4.8 \times 0.5 \times 10^{-12} \,N$
$F = 2.4 \times 10^{-12} \,N$

(D) The magnetic force acting on a moving charge is given by $F = qvB \sin(\theta)$. Since the electron moves in a circular orbit,the velocity is perpendicular to the magnetic field,so $\theta = 90^\circ$ and $\sin(90^\circ) = 1$.
$F = (1.6 \times 10^{-19} \ C) \times (4 \times 10^6 \ m/s) \times (2 \times 10^{-1} \ T) = 1.28 \times 10^{-13} \ N$.
The magnetic force provides the necessary centripetal force for circular motion,so $F = \frac{mv^2}{r}$.
Rearranging for radius $r$: $r = \frac{mv}{qB} = \frac{(9 \times 10^{-31} \ kg) \times (4 \times 10^6 \ m/s)}{1.6 \times 10^{-19} \ C \times 2 \times 10^{-1} \ T}$.
$r = \frac{36 \times 10^{-25}}{3.2 \times 10^{-20}} = 11.25 \times 10^{-5} \ m \approx 1.1 \times 10^{-4} \ m$.

(D) The radius of a circular path for a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.
Since a neutron is electrically neutral $(q = 0)$,it does not experience any magnetic Lorentz force $(F = qvB \sin \theta)$. Therefore,it will not trace a circular path; it will continue to move in a straight line. Thus,Assertion $(A)$ is incorrect.
The period of revolution $T$ is given by $T = \frac{2\pi m}{qB}$. This shows that $T \propto m$. Thus,Reason $(R)$ is correct.
Since $(A)$ is incorrect and $(R)$ is correct,the correct option is $(D)$.

(C) The time period of a charged particle in a uniform magnetic field is given by $T = \frac{2\pi m}{qB}$.
For a proton,$T_p = \frac{2\pi m_p}{eB}$.
For an $\alpha$-particle,$m_{\alpha} = 4m_p$ and $q_{\alpha} = 2e$. Thus,$T_{\alpha} = \frac{2\pi (4m_p)}{(2e)B} = 2 \left( \frac{2\pi m_p}{eB} \right) = 2T_p$.
When the proton's velocity changes direction by $90^{\circ}$,it has completed one-fourth of its circular path,meaning the time elapsed is $t = \frac{T_p}{4}$.
In this same time $t$,the $\alpha$-particle covers an angle $\theta_{\alpha} = \omega_{\alpha} t = \left( \frac{2\pi}{T_{\alpha}} \right) \left( \frac{T_p}{4} \right) = \left( \frac{2\pi}{2T_p} \right) \left( \frac{T_p}{4} \right) = \frac{\pi}{4} = 45^{\circ}$.
Since the particles were projected in opposite directions,let the initial velocity of the proton be $\vec{v}_p = v_0 \hat{i}$ and the $\alpha$-particle be $\vec{v}_{\alpha} = -v_0 \hat{i}$.
After time $t$,the proton's velocity is $\vec{v}_p' = v_0 \hat{j}$ (rotated $90^{\circ}$ counter-clockwise).
The $\alpha$-particle's velocity rotates $45^{\circ}$ clockwise from its initial direction (opposite to the proton's rotation due to opposite charge sign and opposite initial direction),so $\vec{v}_{\alpha}' = v_0 (\cos 45^{\circ} (-\hat{i}) + \sin 45^{\circ} (-\hat{j})) = -v_0 \frac{1}{\sqrt{2}} \hat{i} - v_0 \frac{1}{\sqrt{2}} \hat{j}$.
The angle between $\vec{v}_p'$ and $\vec{v}_{\alpha}'$ is found using the dot product: $\cos \theta = \frac{\vec{v}_p' \cdot \vec{v}_{\alpha}'}{|v_p'||v_{\alpha}'|} = \frac{(0)(-\frac{1}{\sqrt{2}}) + (1)(-\frac{1}{\sqrt{2}})}{(1)(1)} = -\frac{1}{\sqrt{2}}$.
Thus,$\theta = 135^{\circ}$. However,re-evaluating the geometry: the proton turns $90^{\circ}$ and the $\alpha$-particle turns $45^{\circ}$ in the opposite sense relative to the initial axis. The angle between them is $180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ}$.

(C) Given:
Mass of the block, $m = 20 \,g = 0.02 \,kg$
Charge on the block, $q = 4 \,mC = 4 \times 10^{-3} \,C$
Magnetic field, $B = 1 \,T$
Inclination angle, $\theta = 45^{\circ}$
The magnetic force acting on the moving charged block is given by $F_m = qvB$, which acts perpendicular to the inclined plane (upwards).
The block loses contact with the surface when the magnetic force equals the normal component of the gravitational force:
$F_m = mg \cos \theta$
$qvB = mg \cos \theta$
$v = \frac{mg \cos \theta}{qB}$
Since the block is released from rest on a smooth inclined plane, its acceleration is $a = g \sin \theta$. The velocity $v$ at time $t$ is:
$v = u + at = 0 + (g \sin \theta)t = gt \sin \theta$
Equating the two expressions for $v$:
$gt \sin \theta = \frac{mg \cos \theta}{qB}$
$t = \frac{m \cos \theta}{qB \sin \theta} = \frac{m \cot \theta}{qB}$
Substituting the values:
$t = \frac{0.02 \times \cot 45^{\circ}}{4 \times 10^{-3} \times 1}$
$t = \frac{0.02 \times 1}{0.004} = 5 \,s$
Thus, the block loses contact after $5 \,s$.

(C) Since the velocity $v$ is not perpendicular to the magnetic field $B$,the path of the charged particle is a helix.
The pitch of the helical path is given by the formula:
$Pitch = (v \cos \theta) \times T = (v \cos \theta) \times \frac{2 \pi m}{B q}$
Given:
$v = 4 \times 10^5 \ ms^{-1}$
$\theta = 60^{\circ}$
$B = 0.314 \ T$
$m = 1.6 \times 10^{-27} \ kg$
$q = 1.6 \times 10^{-19} \ C$ (charge of a proton)
Substituting the values:
$Pitch = (4 \times 10^5 \times \cos 60^{\circ}) \times \frac{2 \times 3.14 \times 1.6 \times 10^{-27}}{0.314 \times 1.6 \times 10^{-19}}$
$Pitch = (4 \times 10^5 \times 0.5) \times \frac{2 \times 3.14 \times 10^{-27}}{0.314 \times 10^{-19}}$
$Pitch = 2 \times 10^5 \times \frac{6.28 \times 10^{-27}}{0.314 \times 10^{-19}}$
$Pitch = 2 \times 10^5 \times 20 \times 10^{-8}$
$Pitch = 40 \times 10^{-3} \ m = 4 \times 10^{-2} \ m = 4 \ cm$.

(C) The velocity component perpendicular to the magnetic field is $v_{\perp} = v \sin(\theta) = 10^7 \times \sin(30^{\circ}) = 10^7 \times 0.5 = 5 \times 10^6 \,m/s$.
The radius of the circular path is given by $r = \frac{mv_{\perp}}{qB}$,where $m = 9.1 \times 10^{-31} \,kg$ and $q = 1.6 \times 10^{-19} \,C$.
The time period $T$ for one revolution is given by $T = \frac{2\pi r}{v_{\perp}}$.
Substituting the values: $T = \frac{2 \times 3.14 \times 2}{5 \times 10^6} = \frac{12.56}{5 \times 10^6} = 2.512 \times 10^{-6} \,s$.
Rounding to the nearest option,we get $T \approx 2.5 \times 10^{-6} \,s$.

(A) When a charged particle moves perpendicular to a uniform magnetic field,it follows a circular path.
The radius $r$ of the circular path is given by the formula: $r = \frac{mv}{qB}$.
Rearranging for the magnetic field intensity $B$,we get: $B = \frac{mv}{qr}$.
Given values:
Mass $m = 9 \times 10^{-31} \ kg$
Velocity $v = 10^6 \ ms^{-1}$
Charge $q = 1.6 \times 10^{-19} \ C$
Radius $r = 10 \ cm = 0.1 \ m$
Substituting these values into the formula:
$B = \frac{(9 \times 10^{-31}) \times (10^6)}{(1.6 \times 10^{-19}) \times (0.1)}$
$B = \frac{9 \times 10^{-25}}{0.16 \times 10^{-19}}$
$B = \frac{9}{0.16} \times 10^{-6} \ T$
$B = 56.25 \times 10^{-6} \ T = 5.625 \times 10^{-5} \ T$.
Thus,the correct option is $A$.

(A) The pitch of a helical path is given by the formula $p = v \cos(\theta) \times T$,where $T$ is the time period of revolution.
The time period $T$ is given by $T = \frac{2\pi m}{qB}$.
Substituting the values: $m = 1.6 \times 10^{-27} \ kg$,$q = 1.6 \times 10^{-19} \ C$,$v = 1.6 \times 10^5 \ m/s$,$B = \frac{\pi}{10} \ T$,and $\theta = 60^{\circ}$.
$T = \frac{2 \times \pi \times 1.6 \times 10^{-27}}{1.6 \times 10^{-19} \times (\pi / 10)} = \frac{2 \times 10^{-27}}{10^{-19} \times 0.1} = 2 \times 10^{-7} \ s$.
Now,the pitch $p = v \cos(60^{\circ}) \times T = (1.6 \times 10^5) \times (0.5) \times (2 \times 10^{-7}) = 1.6 \times 10^{-2} \ m$.

(B) For an electron to travel in a straight line in a crossed field (where electric and magnetic fields are perpendicular), the net Lorentz force must be zero.
$F_{net} = F_e + F_m = 0$
$eE = evB$
Where $e$ is the electronic charge, $E$ is the electric field intensity, $v$ is the velocity, and $B$ is the magnetic field induction.
$v = \frac{E}{B}$
Given $E = 20 \times 10^3 \,V/m$ and $B = 2.0 \,T$.
$v = \frac{20 \times 10^3}{2.0} = 10 \times 10^3 \,m/s$.

(B) Given: Velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$,Magnetic field $B = (2 \hat{j} + 3 \hat{k}) \ T$,Specific charge $\frac{q}{m} = 0.96 \times 10^8 \ C/kg$.
The force acting on the proton is given by $F = q(v \times B)$.
Calculate the cross product $v \times B$:
$v \times B = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 0 \\ 0 & 2 & 3 \end{vmatrix} = \hat{i}(2 \times 3 - 0 \times 2) - \hat{j}(3 \times 3 - 0 \times 0) + \hat{k}(3 \times 2 - 2 \times 0) = 6 \hat{i} - 9 \hat{j} + 6 \hat{k}$.
Using Newton's second law,$F = ma$,so $a = \frac{F}{m} = \frac{q}{m}(v \times B)$.
Substituting the values:
$a = (0.96 \times 10^8) \times (6 \hat{i} - 9 \hat{j} + 6 \hat{k})$
$a = 0.96 \times 10^8 \times 3 \times (2 \hat{i} - 3 \hat{j} + 2 \hat{k})$
$a = 2.88 \times 10^8 \times 100 \times (2 \hat{i} - 3 \hat{j} + 2 \hat{k})$
$a = 288 \times 10^8(2 \hat{i} - 3 \hat{j} + 2 \hat{k}) \ m/s^2$.

(C) Given: $m = 0.6 \,g = 0.6 \times 10^{-3} \,kg$,$q = 25 \,nC = 25 \times 10^{-9} \,C$,$v = 1.2 \times 10^4 \,ms^{-1}$,$g = 10 \,ms^{-2}$.
Since the particle is moving with a uniform velocity,its acceleration is zero. This implies that the net force acting on the particle is zero.
The magnetic force must balance the gravitational force (weight) acting on the particle.
$F_m = F_g$
$Bqv = mg$
$B = \frac{mg}{qv}$
Substituting the values:
$B = \frac{0.6 \times 10^{-3} \times 10}{25 \times 10^{-9} \times 1.2 \times 10^4}$
$B = \frac{6 \times 10^{-3}}{30 \times 10^{-5}}$
$B = \frac{6 \times 10^2}{30} = \frac{600}{30} = 20 \,T$.

(A) Statement $(A)$ is correct: The magnetic force on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by $\vec{F} = q(\vec{v} \times \vec{B})$. If $\vec{v} = 0$,then $\vec{F} = 0$. Thus,it only exerts force on a moving charge.
Statement $(B)$ is correct: The electric force on a charge $q$ in an electric field $\vec{E}$ is given by $\vec{F} = q\vec{E}$,which acts regardless of whether the charge is stationary or moving.
Statement $(C)$ is incorrect: If the charge moves parallel to the magnetic field,the angle $\theta$ between $\vec{v}$ and $\vec{B}$ is $0^\circ$ or $180^\circ$. Since $\vec{v} \times \vec{B} = vB \sin(\theta)$,the force becomes zero.
Therefore,only statements $(A)$ and $(B)$ are true.

(B) When a charged particle moves in a magnetic field perpendicular to its velocity,the magnetic force $F = q(v \times B)$ acts perpendicular to the velocity vector at every instant.
Since the force is always perpendicular to the velocity,the work done by the magnetic force is $W = F \cdot ds = 0$.
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 as the particle moves in a circular path,the momentum $p = mv$ changes because momentum is a vector quantity.
Therefore,the momentum changes while the kinetic energy remains constant.

(B) The magnetic field due to wire $A$ (along $-\hat{i}$ axis) at point $P(d, d)$ is given by the right-hand rule. The distance of point $P$ from the $x$-axis is $d$. The magnetic field is $B_A = \frac{\mu_0 I}{2 \pi d} (-\hat{k})$.
The magnetic field due to wire $B$ (along $\hat{j}$ axis) at point $P(d, d)$ is given by the right-hand rule. The distance of point $P$ from the $y$-axis is $d$. The magnetic field is $B_B = \frac{\mu_0 I}{2 \pi d} (-\hat{k})$.
The resultant magnetic field at point $P$ is $B = B_A + B_B = \frac{\mu_0 I}{2 \pi d} (-\hat{k}) + \frac{\mu_0 I}{2 \pi d} (-\hat{k}) = \frac{\mu_0 I}{\pi d} (-\hat{k})$.
The force on the charged particle is given by $F = q(v \times B)$.
Given $v = v\hat{i}$ and $B = -\frac{\mu_0 I}{\pi d} \hat{k}$.
$F = q(v\hat{i} \times (-\frac{\mu_0 I}{\pi d} \hat{k})) = -\frac{\mu_0 I q v}{\pi d} (\hat{i} \times \hat{k}) = -\frac{\mu_0 I q v}{\pi d} (-\hat{j}) = \frac{\mu_0 I q v}{\pi d} \hat{j}$.

(A) The kinetic energy of the proton is $K = qV = 1.6 \times 10^{-19} \times 500 \times 10^3 = 8 \times 10^{-14} \ J$.
The momentum $p$ of the proton is given by $p = \sqrt{2mK} = \sqrt{2 \times 1.6 \times 10^{-27} \times 8 \times 10^{-14}} = \sqrt{25.6 \times 10^{-41}} = \sqrt{256 \times 10^{-42}} = 1.6 \times 10^{-20} \ kg \cdot m/s$.
The radius of the circular path in the magnetic field is $R = \frac{p}{qB} = \frac{1.6 \times 10^{-20}}{1.6 \times 10^{-19} \times 0.1} = \frac{1.6 \times 10^{-20}}{1.6 \times 10^{-20}} = 1 \ m$.
For a small thickness $d = 1.0 \ cm = 0.01 \ m$,the angle of deviation $\theta$ is given by $\sin \theta = \frac{d}{R}$.
Since $\theta$ is small,$\sin \theta \approx \theta = \frac{d}{R} = \frac{0.01}{1} = 0.01 \ rad$.

(B) The time period $T$ of a charged particle moving in a uniform magnetic field $B$ is given by the formula $T = \frac{2\pi m}{qB}$,where $m$ is the mass and $q$ is the charge of the particle.
For a proton,$m_p = m$ and $q_p = e$. Thus,$T_p = \frac{2\pi m}{eB}$.
For an alpha particle,$m_{\alpha} = 4m$ and $q_{\alpha} = 2e$. Thus,$T_{\alpha} = \frac{2\pi (4m)}{(2e)B} = \frac{4\pi m}{eB}$.
The ratio of the time taken by the proton to the time taken by the alpha particle is $\frac{T_p}{T_{\alpha}} = \frac{2\pi m / eB}{4\pi m / eB} = \frac{2}{4} = \frac{1}{2}$.
Therefore,the ratio is $1: 2$.

(B) The force on a charged particle moving in a magnetic field is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
Here,the velocity $\vec{v}$ is towards the east (let this be the $+x$ direction) and the magnetic field $\vec{B}$ is directed vertically up (let this be the $+z$ direction).
The direction of the force is given by the cross product of $\vec{v}$ and $\vec{B}$. Using the right-hand rule,$\hat{i} \times \hat{k} = -\hat{j}$.
Thus,the force acts towards the south ($-y$ direction).
Since the force is always perpendicular to the velocity,the particle moves in a circular path. Because the force is horizontal (in the $xy$-plane),the particle will move in a horizontal circular path.
Furthermore,since the magnetic force does no work on the charged particle,its speed remains constant.

(B) The radius $r$ of a charged particle moving in a uniform magnetic field $B$ is given by the formula: $r = \frac{mv}{qB}$.
Rearranging for $B$,we get: $B = \frac{mv}{qr}$.
Given values:
$m = 1.66 \times 10^{-27} \ kg$
$v = 8 \times 10^5 \ m/s$
$q = 1.6 \times 10^{-19} \ C$
$r = 8.3 \ cm = 8.3 \times 10^{-2} \ m$
Substituting these values into the formula:
$B = \frac{(1.66 \times 10^{-27}) \times (8 \times 10^5)}{(1.6 \times 10^{-19}) \times (8.3 \times 10^{-2})}$
$B = \frac{13.28 \times 10^{-22}}{13.28 \times 10^{-21}}$
$B = 0.1 \ T = 100 \ mT$.
Therefore,the magnitude of the magnetic field is $100 \ mT$.

(C) The pitch $p$ of a helical path in a uniform magnetic field is given by the formula: $p = \frac{2 \pi m v \cos \theta}{q B}$.
Since the velocity $v$,the magnetic field $B$,and the angle $\theta$ are the same for both particles,the pitch is proportional to the ratio of mass to charge: $p \propto \frac{m}{q}$.
Therefore,the ratio of the pitches for particles $A$ and $B$ is given by: $\frac{p_A}{p_B} = \frac{m_A / q_A}{m_B / q_B}$.
Given $m_A = m$,$q_A = 2q$,$m_B = 2m$,and $q_B = 3q$.
Substituting these values: $\frac{p_A}{p_B} = \frac{m / 2q}{2m / 3q} = \frac{m}{2q} \times \frac{3q}{2m} = \frac{3}{4}$.
Thus,the ratio is $3:4$.

(A) For an electron falling freely under gravity,its velocity vector is directed downwards,which can be represented as $\overrightarrow{v} = -v_0 \hat{k}$.
The uniform magnetic field is directed towards the south,which can be represented as $\overrightarrow{B} = -B_0 \hat{j}$.
The Lorentz force $\overrightarrow{F}$ on a charged particle is given by $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
For an electron,the charge $q = -e$.
Substituting the values:
$\overrightarrow{F} = -e [(-v_0 \hat{k}) \times (-B_0 \hat{j})]$
$\overrightarrow{F} = -e [v_0 B_0 (\hat{k} \times \hat{j})]$
Since $\hat{k} \times \hat{j} = -\hat{i}$,we have:
$\overrightarrow{F} = -e [v_0 B_0 (-\hat{i})]$
$\overrightarrow{F} = e v_0 B_0 \hat{i}$
The direction $\hat{i}$ corresponds to the east direction.
Therefore,the electron is initially deflected towards the east.

(A) The velocity of the charge is $\vec{v} = 2 \hat{i} \ m/s$.
The magnetic field is $\vec{B} = (2 \hat{i} + 2 \hat{j} + 3 \hat{k}) \ T$.
The magnetic force $\vec{F}$ on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Substituting the values:
$\vec{F} = q(2 \hat{i}) \times (2 \hat{i} + 2 \hat{j} + 3 \hat{k})$
$\vec{F} = q [ (2 \hat{i} \times 2 \hat{i}) + (2 \hat{i} \times 2 \hat{j}) + (2 \hat{i} \times 3 \hat{k}) ]$
Since $\hat{i} \times \hat{i} = 0$,$\hat{i} \times \hat{j} = \hat{k}$,and $\hat{i} \times \hat{k} = -\hat{j}$:
$\vec{F} = q [ 0 + 4 \hat{k} - 6 \hat{j} ]$
$\vec{F} = q(4 \hat{k} - 6 \hat{j})$.
This force vector lies in the $y-z$ plane.

(A) According to the Lorentz force law,the magnetic force $F$ acting on a charged particle with charge $q$ moving with velocity $v$ in a magnetic field $B$ is given by the vector product:
$F = q(v \times B)$
Given that the particle is an electron,its charge is $q = -e$ (where $e$ is the magnitude of the elementary charge).
Substituting $q = -e$ into the formula,we get:
$F = -e(v \times B)$
However,in the context of standard multiple-choice questions where the magnitude or the form of the Lorentz force is requested,the expression $e(v \times B)$ represents the vector form of the force acting on a charge $e$ moving in a magnetic field.
Thus,the correct option is $A$.

(B) Initially,the charged particle moves along the magnetic field $B$,so the magnetic force is zero.
When a uniform electric field $E$ is applied perpendicular to $B$,the particle experiences an electric force $F = qE$ in the direction of $E$.
This force accelerates the particle,giving it a velocity component perpendicular to $B$.
As the particle now has velocity components both parallel and perpendicular to the magnetic field $B$,it follows a helical path.
Specifically,in the presence of crossed electric and magnetic fields,the particle undergoes cycloidal motion if the initial velocity is zero,but with an initial velocity along $B$,the trajectory is a helix.

(D) When a charged particle with charge $q$ and mass $m$ enters a uniform magnetic field $B$ with velocity $v$ perpendicular to the field,it experiences a magnetic Lorentz force $F = q(v \times B)$.
Since the force is always perpendicular to the velocity,it acts as a centripetal force,causing the particle to move in a circular path.
The radius $r$ of this circular path is given by the formula:
$r = \frac{mv}{qB}$
From this expression,it is clear that the radius $r$ is directly proportional to the velocity $v$ $(r \propto v)$.
Therefore,the particle travels in a circular path with a radius directly proportional to its velocity.

(C) The kinetic energy of the electron is given by $K = \frac{1}{2}mv^2$.
Given $K = 45.5 \ eV = 45.5 \times 1.6 \times 10^{-19} \ J$ and $m = 9.1 \times 10^{-31} \ kg$.
$v^2 = \frac{2K}{m} = \frac{2 \times 45.5 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}} = \frac{145.6 \times 10^{-19}}{9.1 \times 10^{-31}} = 16 \times 10^{12} \ m^2s^{-2}$.
Thus,$v = 4 \times 10^6 \ ms^{-1}$.
For an electron to remain undeflected in crossed electric and magnetic fields,the electric force must balance the magnetic force: $eE = evB$,which implies $v = \frac{E}{B}$.
Given $E = 1 \times 10^3 \ Vm^{-1}$,we have $B = \frac{E}{v} = \frac{1 \times 10^3}{4 \times 10^6} = 0.25 \times 10^{-3} = 2.5 \times 10^{-4} \ Wb \ m^{-2}$.

(C) When a charged particle moves perpendicularly to a uniform magnetic field,it follows a circular path.
The radius $r$ of the circular path is given by the formula:
$r = \frac{mv}{qB}$
Given values:
Mass of electron $m = 9 \times 10^{-31} \ kg$
Velocity $v = 1.6 \times 10^7 \ m/s$
Charge of electron $e = 1.6 \times 10^{-19} \ C$
Magnetic field $B = 0.1 \ T$
Substituting these values into the formula:
$r = \frac{9 \times 10^{-31} \times 1.6 \times 10^7}{1.6 \times 10^{-19} \times 0.1}$
$r = \frac{9 \times 1.6 \times 10^{-24}}{1.6 \times 10^{-20}}$
$r = 9 \times 10^{-4} \ m$

(C) When a charged particle moves perpendicular to a uniform magnetic field,it follows a circular path. On a circular path,the velocity vector is always tangent to the path,and the acceleration vector (centripetal acceleration) is directed towards the center of the circle. Therefore,the velocity and acceleration vectors are always perpendicular to each other at any instant.
Given,$\overrightarrow{v}=2 \hat{i}+c \hat{j}$ and $\overrightarrow{a}=3 \hat{i}+4 \hat{j}$.
Since $\overrightarrow{v} \perp \overrightarrow{a}$,their dot product must be zero:
$\overrightarrow{v} \cdot \overrightarrow{a} = 0$
$(2 \hat{i} + c \hat{j}) \cdot (3 \hat{i} + 4 \hat{j}) = 0$
$(2)(3) + (c)(4) = 0$
$6 + 4c = 0$
$4c = -6$
$c = -1.5$

(B) The radius of a circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.
Since kinetic energy $K$ and magnetic field $B$ are the same for all particles, we have $r \propto \frac{\sqrt{m}}{q}$.
For a proton $(p)$: mass $m_p = m$, charge $q_p = q$. So, $r_p \propto \frac{\sqrt{m}}{q}$.
For a deuteron $(d)$: mass $m_d = 2m$, charge $q_d = q$. So, $r_d \propto \frac{\sqrt{2m}}{q}$.
For an $\alpha$-particle $(\alpha)$: mass $m_\alpha = 4m$, charge $q_\alpha = 2q$. So, $r_\alpha \propto \frac{\sqrt{4m}}{2q} = \frac{2\sqrt{m}}{2q} = \frac{\sqrt{m}}{q}$.
Therefore, the ratio $r_p : r_d : r_\alpha = \frac{\sqrt{m}}{q} : \frac{\sqrt{2m}}{q} : \frac{\sqrt{m}}{q} = 1 : \sqrt{2} : 1$.

(D) The force on a charged particle moving in a magnetic field is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
Case $(1)$: If the velocity $\vec{v}$ is parallel or anti-parallel to the magnetic field $\vec{B}$,then $\vec{v} \times \vec{B} = 0$. The force is zero,and the particle continues in a straight line.
Case $(2)$: If the velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,the magnetic force acts as a centripetal force,causing the particle to move in a circle.
Case $(3)$: If the velocity $\vec{v}$ makes an angle $\theta$ (where $\theta \neq 0^\circ, 90^\circ, 180^\circ$) with the magnetic field $\vec{B}$,the velocity can be resolved into two components: one parallel to $\vec{B}$ (causing linear motion) and one perpendicular to $\vec{B}$ (causing circular motion). The resultant trajectory is a helix.
Therefore,all three trajectories are possible depending on the angle between the velocity and the magnetic field.

(A) Given ratios: $m_1: m_2 = 1: 1$,$q_1: q_2 = 1: 2$,and $v_1: v_2 = 2: 3$.
The radius $r$ of the circular path of a charged particle moving in a uniform magnetic field $B$ is given by the formula $r = \frac{mv}{Bq}$.
Taking the ratio of the radii for the two particles,we get:
$\frac{r_1}{r_2} = \left(\frac{m_1}{m_2}\right) \times \left(\frac{v_1}{v_2}\right) \times \left(\frac{q_2}{q_1}\right)$.
Substituting the given values:
$\frac{r_1}{r_2} = \left(\frac{1}{1}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{2}{1}\right) = \frac{4}{3}$.
Therefore,the ratio of the radii is $4: 3$.

(B) Given: Velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$,Magnetic field $\vec{B} = (2 \hat{j} + 3 \hat{k}) \text{ T}$.
Specific charge $\frac{q}{m} = 0.96 \times 10^8 \text{ C kg}^{-1}$.
The magnetic force on a moving charge is $\vec{F} = q(\vec{v} \times \vec{B})$.
According to Newton's second law,$\vec{F} = m\vec{a}$,so $\vec{a} = \frac{q}{m}(\vec{v} \times \vec{B})$.
First,calculate the cross product $\vec{v} \times \vec{B}$:
$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 0 \\ 0 & 2 & 3 \end{vmatrix} = \hat{i}(2 \times 3 - 0 \times 2) - \hat{j}(3 \times 3 - 0 \times 0) + \hat{k}(3 \times 2 - 2 \times 0) = 6 \hat{i} - 9 \hat{j} + 6 \hat{k}$.
Now,calculate acceleration $\vec{a} = \frac{q}{m}(6 \hat{i} - 9 \hat{j} + 6 \hat{k})$:
$\vec{a} = 0.96 \times 10^8 \times (6 \hat{i} - 9 \hat{j} + 6 \hat{k})$
$\vec{a} = 0.96 \times 10^8 \times 3 \times (2 \hat{i} - 3 \hat{j} + 2 \hat{k})$
$\vec{a} = 2.88 \times 10^8 \times (2 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 288 \times 10^6 \times (2 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 288 \times 10^8 \times (2 \hat{i} - 3 \hat{j} + 2 \hat{k}) \text{ ms}^{-2}$ (Note: The numerical coefficient matches option $B$).

(C) The velocity of the electron is $\vec{v} = 2 \times 10^5 \hat{i} \ m/s$.
The magnetic field is $\vec{B} = (\hat{i} + 4\hat{j} - 3\hat{k}) \ T$.
The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
For an electron,$q = -1.6 \times 10^{-19} \ C$.
Calculating the cross product $\vec{v} \times \vec{B}$:
$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 \times 10^5 & 0 & 0 \\ 1 & 4 & -3 \end{vmatrix} = \hat{i}(0) - \hat{j}(-6 \times 10^5) + \hat{k}(8 \times 10^5) = (6 \times 10^5 \hat{j} + 8 \times 10^5 \hat{k}) \ m/s \cdot T$.
The magnitude of the cross product is $|\vec{v} \times \vec{B}| = \sqrt{(6 \times 10^5)^2 + (8 \times 10^5)^2} = \sqrt{36 \times 10^{10} + 64 \times 10^{10}} = \sqrt{100 \times 10^{10}} = 10 \times 10^5 = 10^6 \ m/s \cdot T$.
The magnitude of the force is $F = |q| |\vec{v} \times \vec{B}| = (1.6 \times 10^{-19} \ C) \times (10^6 \ m/s \cdot T) = 1.6 \times 10^{-13} \ N$.

(C) Given: Mass $m = 0.6 \,g = 0.6 \times 10^{-3} \,kg$, Charge $q = 25 \,nC = 25 \times 10^{-9} \,C$, Velocity $v = 1.2 \times 10^4 \,ms^{-1}$, Acceleration due to gravity $g = 10 \,ms^{-2}$.
Since the particle moves with a uniform velocity, the net force acting on it must be zero.
This implies that the magnetic force $F_m$ must balance the gravitational force $F_g$ acting on the particle.
$F_m = F_g$
$Bqv = mg$
$B = \frac{mg}{qv}$
Substituting the values:
$B = \frac{0.6 \times 10^{-3} \times 10}{25 \times 10^{-9} \times 1.2 \times 10^4}$
$B = \frac{6 \times 10^{-3}}{30 \times 10^{-5}}$
$B = \frac{6 \times 10^2}{30} = \frac{600}{30} = 20 \,T$
Therefore, the magnetic induction is $20 \,T$.

(B) The magnetic force is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Since $\vec{F}$ is perpendicular to $\vec{v}$,their dot product must be zero: $\vec{F} \cdot \vec{v} = 0$.
$(F_1 \hat{i} + F_2 \hat{j}) \cdot (v_1 \hat{i} + v_2 \hat{j}) = F_1 v_1 + F_2 v_2 = 0 \Rightarrow \frac{F_1}{F_2} = -\frac{v_2}{v_1} \quad (I)$.
Also,$\vec{F}$ is perpendicular to $\vec{B}$,so $\vec{F} \cdot \vec{B} = 0$.
Let $\vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}$. Then $(F_1 \hat{i} + F_2 \hat{j}) \cdot (B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}) = F_1 B_1 + F_2 B_2 = 0 \Rightarrow \frac{F_1}{F_2} = -\frac{B_2}{B_1} \quad (II)$.
Comparing $(I)$ and $(II)$,we get $\frac{v_2}{v_1} = \frac{B_2}{B_1}$,which implies $\frac{v_1}{v_2} = \frac{B_1}{B_2}$.
Thus,$\vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k}$ satisfies the condition $\frac{v_1}{v_2} = \frac{B_1}{B_2}$.

(B) The magnetic field is directed along the $z$-axis $(B = B_0 \hat{k})$.
The velocity of the particle is $v = 3 \hat{i} + 4 \hat{k} \text{ m/s}$.
The velocity component perpendicular to the magnetic field is $v_{\perp} = 3 \hat{i} \text{ m/s}$,which causes circular motion in the $x-y$ plane.
The velocity component parallel to the magnetic field is $v_{\parallel} = 4 \hat{k} \text{ m/s}$,which remains constant as there is no force acting in this direction.
Since the particle has both perpendicular and parallel velocity components,the trajectory is a helical path.
The distance to be covered along the $z$-axis is $s = 2 \text{ m}$.
Using the formula $s = v_{\parallel} \times t$,we get $t = \frac{s}{v_{\parallel}} = \frac{2}{4} = \frac{1}{2} \text{ s}$.

(A, B, C, D) For a charged particle in a magnetic field,the radius of the circular path is given by $r = \frac{mv}{qB}$.
Applying Bohr's quantization condition for angular momentum: $mvr = \frac{nh}{2\pi} = n\hbar$.
Substituting $v = \frac{qBr}{m}$ into the quantization condition: $m \left( \frac{qBr}{m} \right) r = n\hbar \implies qBr^2 = n\hbar \implies r_n = \sqrt{\frac{n\hbar}{qB}}$. Thus,$r_n \propto \sqrt{n}$. (Statement $A$ is true).
Now,$v_n = \frac{qBr_n}{m} = \frac{qB}{m} \sqrt{\frac{n\hbar}{qB}} = \sqrt{\frac{n q B \hbar}{m^2}}$. (Statement $B$ is true).
The energy of the $n^{\text{th}}$ level is $E_n = \frac{1}{2}mv_n^2 = \frac{1}{2}m \left( \frac{n q B \hbar}{m^2} \right) = n \left( \frac{q B \hbar}{2m} \right)$. Thus,$E_n \propto n$. (Statement $C$ is true).
The transition frequency $\omega$ is given by $\Delta E = \hbar \omega$. For successive levels,$\Delta E = E_{n+1} - E_n = \frac{q B \hbar}{2m}$. Therefore,$\omega = \frac{\Delta E}{\hbar} = \frac{qB}{2m}$,which is independent of $n$. (Statement $D$ is true).