Motion of Charged Particle In Magnetic Field Questions in English

(D) $1$. According to the right-hand thumb rule,the magnetic field $\vec{B}$ produced by the current-carrying wire at point $P$ is directed perpendicular to the plane of the paper and inwards (into the page).
$2$. The force $\vec{F}$ on a moving charge $+Q$ in a magnetic field is given by the Lorentz force formula: $\vec{F} = Q(\vec{V} \times \vec{B})$.
$3$. Here,the velocity $\vec{V}$ is along the positive $X$-axis (to the right) and the magnetic field $\vec{B}$ is directed inwards (into the page,represented by $-\hat{k}$ direction).
$4$. Using the right-hand rule for the cross product $\vec{V} \times \vec{B}$,where $\vec{V}$ is along $\hat{i}$ and $\vec{B}$ is along $-\hat{k}$,the direction of the force is $\hat{i} \times (-\hat{k}) = -(\hat{i} \times \hat{k}) = -(-\hat{j}) = \hat{j}$.
$5$. This corresponds to the positive $Y$-axis direction,which is along $OY$.

(C) 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 velocity $\vec{v}$ of the electron is directed from south to north.
The magnetic field $\vec{B}$ is directed vertically downwards.
According to the right-hand rule for the cross product $\vec{v} \times \vec{B}$,the direction of $\vec{v} \times \vec{B}$ is towards the west.
Since the electron has a negative charge $(q = -e)$,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 east.
Alternatively,using Fleming's Left-Hand Rule for a negative charge,the force is directed towards the east.

(C) For an electron to move in a straight line through crossed electric and magnetic fields,the net Lorentz force must be zero.
This occurs when the electric force $F_E = qE$ is balanced by the magnetic force $F_B = qvB$.
Setting $F_E = F_B$,we get $qE = qvB$.
Solving for the velocity $v$,we have $v = \frac{E}{B}$.
Given $E = 1500\, V/m$ and $B = 0.40\, Wb/m^2$,the speed is $v = \frac{1500}{0.40} = 3750\, m/s$.
This can be expressed in scientific notation as $3.75 \times 10^3\, m/s$.

(C) When a charged particle moves in a uniform magnetic field,the magnetic Lorentz force provides the necessary centripetal force for circular motion.
$F_m = F_c$
$qvB = \frac{mv^2}{r}$
Rearranging for the radius $r$:
$r = \frac{mv}{qB}$
Since $m$,$q$,and $B$ are constants for a given electron in a uniform magnetic field,we have:
$r \propto v$
Therefore,the radius is directly proportional to the speed of the electron.

(B) $1$. According to the right-hand thumb rule,the magnetic field produced by the current-carrying conductor at the position of the electron stream (below the conductor) is directed into the plane of the paper.
$2$. The electrons are moving to the right. The force on a moving charge $q$ in a magnetic field $B$ is given by the Lorentz force formula $F = q(v \times B)$.
$3$. Since the charge of an electron is negative $(q = -e)$,the direction of the force is opposite to the direction of $(v \times B)$.
$4$. Using the right-hand rule for the cross product $(v \times B)$,where $v$ is to the right and $B$ is into the page,the direction of $(v \times B)$ is upward.
$5$. Therefore,the force on the electrons is downward. Alternatively,using Fleming's left-hand rule for negative charges,the force is directed downward.

(D) 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 $v$ into the radius formula: $r = \frac{m}{qB} \sqrt{\frac{2K}{m}} = \frac{\sqrt{2mK}}{qB}$.
Since $K$ and $B$ are constant,$r \propto \frac{\sqrt{m}}{q}$.
Given mass ratios $m_H:m_{He}:m_O = 1:4:16$ and charges $q_H=1, q_{He}=1, q_O=2$ (in units of $e$):
$r_H \propto \frac{\sqrt{1}}{1} = 1$
$r_{He} \propto \frac{\sqrt{4}}{1} = 2$
$r_O \propto \frac{\sqrt{16}}{2} = \frac{4}{2} = 2$
Thus,the ratio of radii is $r_H:r_{He}:r_O = 1:2:2$.
Since deflection is inversely proportional to the radius $(d \propto 1/r)$,the ion with the smallest radius $(H^+)$ is deflected the most,and $He^+$ and $O^{++}$ have the same radius,hence the same deflection. Therefore,both $(a)$ and $(c)$ are correct.

(C) The electric field $\vec{E}$ is along the $+x$ direction. Positive ions experience an electric force $\vec{F}_e = q\vec{E}$ along the $+x$ direction,while negative ions experience an electric force along the $-x$ direction.
As a result,positive ions acquire a velocity $\vec{v}$ in the $+x$ direction,and negative ions acquire a velocity $\vec{v}$ in the $-x$ direction.
The magnetic field $\vec{B}$ is along the $+z$ direction.
The magnetic force on a charged particle is given by $\vec{F}_m = q(\vec{v} \times \vec{B})$.
For positive ions: $\vec{F}_m = (+q)(v\hat{i} \times B\hat{k}) = -qvB\hat{j}$,which is in the $-y$ direction.
For negative ions: $\vec{F}_m = (-q)(-v\hat{i} \times B\hat{k}) = (-q)(-vB(-\hat{j})) = -qvB\hat{j}$,which is also in the $-y$ direction.
Thus,both types of ions deflect towards the $-y$ direction.

(B) The particle moves in a straight line until it reaches $x = a$. Upon entering the magnetic field region $(a \le x \le b)$,the magnetic force acts as a centripetal force,causing the particle to move in a circular path of radius $r = \frac{mv}{qB}$.
For the particle to just reach the region $x > b$,the radius of the circular path must be at least equal to the width of the magnetic field region,which is $(b - a)$.
Therefore,the condition is $r \ge (b - a)$.
Substituting the expression for $r$,we get $\frac{mv}{qB} \ge (b - a)$.
Solving for $v$,we find $v \ge \frac{q(b - a)B}{m}$.
Thus,the minimum velocity required is $v_{\min} = \frac{q(b - a)B}{m}$.

(B) The particle moves in the $x-y$ plane. For the motion to remain in the $x-y$ plane,the net force $\overrightarrow F_{net} = q(\overrightarrow E + \overrightarrow v \times \overrightarrow B)$ must have no component along the $z$-axis.
Initial velocity is $\overrightarrow v = v\hat i$.
In option $(d)$,$\overrightarrow F_{net} = q(a\hat i) + q(v\hat i \times (c\hat k + b\hat j)) = qa\hat i - qvc\hat j + qvb\hat k$. The presence of a $z$-component $(qvb\hat k)$ forces the particle out of the $x-y$ plane,so $(d)$ is incorrect.
In option $(b)$,$\overrightarrow F_{net} = q(a\hat i) + q(v\hat i \times (c\hat k + a\hat i)) = qa\hat i - qvc\hat j$. Here,the force has only $x$ and $y$ components,keeping the particle in the $x-y$ plane. The path is non-circular because the electric field changes the speed of the particle. Thus,option $(b)$ is the correct combination.

(A) Since the block is metallic,the charge carriers are electrons. For a current $I$ along the positive $x$-axis,the electrons move along the negative $x$-axis,i.e.,$\overrightarrow{v} = -v\hat{i}$.
The magnetic field is along the $y$-axis,i.e.,$\overrightarrow{B} = B\hat{j}$.
The Lorentz force is given by $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$.
Substituting the values,$\overrightarrow{F} = (-e)(-v\hat{i} \times B\hat{j}) = evB(\hat{i} \times \hat{j}) = evB\hat{k}$.
Since the force on the electrons is in the positive $z$-direction,they accumulate on the face $ABCD$ (as per the coordinate system shown in the figure). The accumulation of negative charges on face $ABCD$ results in the lowering of its potential.

(D) Let the two wires be placed in the $xy$-plane,parallel to the $y$-axis,at $x = -d/2$ and $x = d/2$. The currents are in opposite directions (e.g.,$+I$ in the $y$-direction and $-I$ in the $y$-direction).
At the midpoint (origin),the magnetic field $B_1$ due to the first wire is directed into the plane ($-z$ direction),and the magnetic field $B_2$ due to the second wire is also directed into the plane ($-z$ direction).
The resultant magnetic field $B = B_1 + B_2$ is perpendicular to the plane of the wires (along the $-z$ direction).
The velocity $v$ of the charge $q$ is given as perpendicular to the plane of the wires (along the $z$ direction).
The magnetic force on a moving charge is given by $F = q(v \times B)$.
Since the velocity vector $v$ and the magnetic field vector $B$ are collinear (both along the $z$-axis),the cross product $v \times B = 0$.
Therefore,the magnitude of the magnetic force acting on the charge is $0$.

(A) When a charged particle enters a uniform magnetic field perpendicularly,it follows a circular path of radius $r = \frac{mv}{qB} = \frac{p}{qB}$.
From the geometry of the path,the particle travels a horizontal distance $d$ within the magnetic field region.
In the right-angled triangle formed by the radius $r$,the horizontal distance $d$,and the deflection angle $\theta$,we have $\sin \theta = \frac{d}{r}$.
Substituting the value of $r$,we get $\sin \theta = \frac{d}{p / (qB)} = \frac{Bqd}{p}$.
Thus,the correct option is $A$.

(B) From the geometry of the path of the proton in the magnetic field,we have $\sin \theta = \frac{d}{r}$,where $d = 0.1 \; m$ is the width of the magnetic field region and $r$ is the radius of the circular path.
The radius of the circular path is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$,where $V = 500 \times 10^3 \; V$ is the accelerating potential.
Substituting $r$ into the expression for $\sin \theta$:
$\sin \theta = \frac{d}{r} = Bd \sqrt{\frac{q}{2mV}}$
Given values: $B = 0.51 \; T$,$d = 0.1 \; m$,$q = 1.6 \times 10^{-19} \; C$,$m = 1.67 \times 10^{-27} \; kg$,$V = 500 \times 10^3 \; V$.
$\sin \theta = 0.51 \times 0.1 \times \sqrt{\frac{1.6 \times 10^{-19}}{2 \times 1.67 \times 10^{-27} \times 500 \times 10^3}}$
$\sin \theta = 0.051 \times \sqrt{\frac{1.6 \times 10^{-19}}{1.67 \times 10^{-21}}} \approx 0.051 \times \sqrt{95.8} \approx 0.051 \times 9.78 \approx 0.5$
Since $\sin \theta = 0.5$,we have $\theta = 30^o$.

(A) The force on a moving charge in a magnetic field is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
For the electron to reverse its direction,it must follow a semi-circular path in a plane perpendicular to the magnetic field.
If the electron is moving along the positive $X$-axis (velocity $\vec{v} = v\hat{i}$),and we apply a magnetic field along the $Y$-axis $(\vec{B} = B\hat{j})$,the force will be $\vec{F} = -e(v\hat{i} \times B\hat{j}) = -evB\hat{k}$.
This force acts in the $Z$-direction,causing the electron to move in a semi-circle in the $X-Z$ plane.
After completing a semi-circle,the velocity vector will point in the negative $X$-direction $(-v\hat{i})$,effectively reversing the electron's direction.
Thus,applying the magnetic field along the $Y$-axis (or $Z$-axis) will achieve the desired result.

(D) The magnetic force acting on a charged particle is given by $\overrightarrow F = q(\overrightarrow v \times \overrightarrow B)$.
Since the magnetic force is always perpendicular to the velocity of the particle,it does no work on the electron. Therefore,the kinetic energy of the electron remains constant,which implies that the speed $v$ of the electron remains equal to its initial speed $u$,so $v = u$.
Using the right-hand rule for the magnetic force on a negative charge (electron): the velocity $\overrightarrow v$ is along the $+x$-direction and the magnetic field $\overrightarrow B$ is along the $-z$-direction $(-B_0 \hat k)$. The force $\overrightarrow F = -e(\overrightarrow v \times \overrightarrow B) = -e(u \hat i \times -B_0 \hat k) = -e(u B_0 \hat j) = -e u B_0 \hat j$. Thus,the initial force is in the negative $y$-direction.
The electron will follow a circular path in the clockwise direction and will exit the magnetic field region at a point where $y < 0$.

(C) For an electron to pass through the region undeviated,the net Lorentz force acting on it must be zero. The total force is given by $\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B})$.
For $\overrightarrow{F} = 0$,we must have $\overrightarrow{F_e} = -\overrightarrow{F_m}$,which implies $|\overrightarrow{F_e}| = |\overrightarrow{F_m}|$.
This condition is satisfied when the electric force $\overrightarrow{F_e} = q\overrightarrow{E}$,the magnetic force $\overrightarrow{F_m} = q(\overrightarrow{v} \times \overrightarrow{B})$,and the velocity $\overrightarrow{v}$ are oriented such that $\overrightarrow{v}$,$\overrightarrow{E}$,and $\overrightarrow{B}$ are mutually perpendicular to each other.
In option $C$,the velocity $\overrightarrow{v}$,electric field $\overrightarrow{E}$,and magnetic field $\overrightarrow{B}$ are shown to be mutually perpendicular (each at $90^\circ$ to the others),which allows the electric and magnetic forces to balance each other.

(D) The magnetic force acting on a charged particle moving in a magnetic field 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 electron is zero $(W = \vec{F} \cdot \vec{d} = 0)$.
Because the work done is zero,the kinetic energy of the electron remains constant.
Since the kinetic energy $(K = \frac{1}{2}mv^2)$ is constant and the mass $m$ is constant,the speed $v$ remains constant.
Consequently,the magnitude of the momentum $(p = mv)$ also remains unchanged.
Therefore,both energy and momentum remain unchanged.

(C) When an electron beam passes undeviated through mutually perpendicular electric and magnetic fields,the electric force is balanced by the magnetic force.
$F_e = F_m$
$eE = evB$
Therefore,the speed $v$ is given by:
$v = \frac{E}{B}$
Substituting the given values:
$v = \frac{3 \times 10^4}{2 \times 10^{-3}}$
$v = 1.5 \times 10^7 \ m/s$
Thus,the correct option is $C$.

(B) When a charged particle moves with a constant velocity through a region containing both electric field $(E)$ and magnetic field $(B)$ that are mutually perpendicular to each other and to the velocity vector,the net Lorentz force acting on the particle is zero.
Given,the electric field $E = 20 \ V m^{-1}$ and the magnetic field $B = 0.5 \ T$.
The condition for the particle to move with constant velocity is given by the balance of electric and magnetic forces:
$qE = qvB$
$v = \frac{E}{B}$
Substituting the given values:
$v = \frac{20}{0.5} = 40 \ m s^{-1}$.
Thus,the velocity of the electrons is $40 \ m s^{-1}$.

(C) In the Thomson experiment,the condition for no deflection of a charged particle moving through crossed electric field $E$ and magnetic field $B$ is given by the balance of forces: $qE = qvB$,which implies $v = E/B$.
The kinetic energy gained by the particle accelerated through a potential difference $V$ is $qV = \frac{1}{2}mv^2$,which gives $v = \sqrt{\frac{2qV}{m}}$.
Equating the two expressions for velocity: $\frac{E}{B} = \sqrt{\frac{2qV}{m}}$.
Squaring both sides: $\frac{E^2}{B^2} = \frac{2qV}{m}$,which leads to $\frac{q}{m} = \frac{E^2}{2VB^2}$.
For a fixed potential $V$ and electric field $E$,we have $B^2 \propto \frac{1}{m}$.
If the mass $m$ is increased by a factor of $208$,then $B^2$ must change by a factor of $1/208$.
Therefore,$B$ must change by a factor of $\sqrt{1/208}$,meaning $B$ must be multiplied by $\sqrt{208} \approx 14.4$ to maintain the condition.

(C) In a velocity selector,the electric force $F_e = qE$ and the magnetic force $F_m = qvB$ act in opposite directions.
For the ions to pass undeflected,the forces must be balanced: $qE = qvB$,which simplifies to $v = \frac{E}{B}$.
The electric field $E$ between the plates is given by $E = \frac{V}{d}$.
Given $V = 1000 \ V$ and $d = 1 \ cm = 0.01 \ m = 10^{-2} \ m$.
$E = \frac{1000}{10^{-2}} = 10^5 \ V/m$.
Given $B = 1 \ T$.
Substituting these values into the velocity formula: $v = \frac{10^5}{1} = 10^5 \ m/s$.

(B) In Thomson's positive ray experiment,the deflection of ions in electric and magnetic fields is governed by the Lorentz force and kinematics.
The displacement $y$ along the electric field direction is proportional to $(q/m) \cdot (1/v^2)$,and the displacement $z$ along the magnetic field direction is proportional to $(q/m) \cdot (1/v)$.
By eliminating the velocity $v$,we obtain the trajectory equation: $z^2 = k \cdot (q/m) \cdot y$,where $k$ is a constant related to the field strengths and geometry.
Since $q/m$ is constant for all ions,the equation simplifies to the form $z^2 = C \cdot y$,which represents a parabola.
Therefore,the trace on the photographic plate is parabolic.

(B) In a mass spectrograph,when ions are projected with the same velocity $v$ in a uniform magnetic field $B$,they follow a circular path of radius $r$ given by $r = \frac{mv}{qB}$.
Since the ions strike a photographic plate,the distance from the point of entry is the diameter $D = 2r = \frac{2mv}{qB}$.
Given that $v$ and $B$ are constant,the diameter $D$ is directly proportional to the mass-to-charge ratio: $D \propto \frac{m}{q}$.
Calculating the $\frac{m}{q}$ ratio for each ion (taking mass number $A$ as proportional to mass $m$):
$1$. For ${O^{++}}$: $\frac{m}{q} \propto \frac{16}{2} = 8$
$2$. For ${C^+}$: $\frac{m}{q} \propto \frac{12}{1} = 12$
$3$. For ${He^{++}}$: $\frac{m}{q} \propto \frac{4}{2} = 2$
$4$. For ${H^+}$: $\frac{m}{q} \propto \frac{1}{1} = 1$
Since ${C^+}$ has the highest $\frac{m}{q}$ ratio,it will strike the farthest.

(B) For an electron beam to pass undeflected through crossed electric and magnetic fields,the electric force must be equal and opposite to the magnetic force.
$F_e = F_m$
$qE = qvB$
$v = \frac{E}{B}$
Given $E = 1.125 \times 10^{-6} \, N/C$ and $B = 3 \times 10^{-10} \, T$.
$v = \frac{1.125 \times 10^{-6}}{3 \times 10^{-10}} = 0.375 \times 10^4 = 3750 \, m/s$.

(A) In a parabola spectrograph,the deflection $x$ of an ion is inversely proportional to the square of its velocity,i.e.,$x \propto \frac{1}{v^2}$.
This means that the ion which experiences less deflection will have a higher velocity.
From the given graph,the horizontal distances (deflections) of the ions $P, Q, R,$ and $S$ from the vertical axis are $x_1, x_2, x_3,$ and $x_4$ respectively.
Observing the graph,we can see that $x_1 < x_2 < x_3 < x_4$.
Since $v \propto \frac{1}{\sqrt{x}}$,it follows that $v_1 > v_2 > v_3 > v_4$.

(D) In this region,the charged particle experiences an electric force given by $\vec{F_e} = q\vec{E}$.
Since the particle starts from rest,it will begin to accelerate in the direction of the electric field (if positive) or opposite to it (if negative).
As the particle gains velocity $\vec{v}$,it experiences a magnetic force given by $\vec{F_m} = q(\vec{v} \times \vec{B})$.
Since the electric and magnetic fields are parallel,and the particle's velocity $\vec{v}$ is generated by the electric field,the velocity $\vec{v}$ will always be parallel to the magnetic field $\vec{B}$ (i.e.,$\vec{v} \parallel \vec{B}$).
Therefore,the magnetic force $F_m = qvB \sin(0^\circ) = 0$.
Since there is no magnetic force acting on the particle and the electric force acts along the line of the fields,the particle will move in a straight line.

(B) In $J$.$J$. Thomson's experiment to determine the specific charge $(e/m)$ of an electron,the electron beam is subjected to both electric and magnetic fields.
$1$. The electric field $(E)$ and the magnetic field $(B)$ are applied perpendicular to each other.
$2$. Both the electric field $(E)$ and the magnetic field $(B)$ are applied perpendicular to the direction of the electron beam.
$3$. By adjusting the fields such that the electron beam passes undeflected,the velocity of the electrons can be determined,and subsequently,the $e/m$ ratio is calculated.
Therefore,the electric and magnetic fields are perpendicular to each other and both are perpendicular to the electron beam.

(C) The time period of the proton $(T_p)$ is the time taken for one revolution.
Given that the proton takes $25 \mu \text{sec}$ for $5$ revolutions,$T_p = \frac{25}{5} = 5 \mu \text{sec}$.
The formula for the time period of a charged particle in a magnetic field is $T = \frac{2\pi m}{qB}$.
Comparing the time periods of the $\alpha$-particle $(T_{\alpha})$ and the proton $(T_p)$: $\frac{T_{\alpha}}{T_p} = \frac{m_{\alpha}}{m_p} \times \frac{q_p}{q_{\alpha}}$.
Since $m_{\alpha} = 4m_p$ and $q_{\alpha} = 2q_p$,we have $\frac{T_{\alpha}}{T_p} = \frac{4m_p}{m_p} \times \frac{q_p}{2q_p} = 4 \times \frac{1}{2} = 2$.
Therefore,$T_{\alpha} = 2 \times T_p = 2 \times 5 \mu \text{sec} = 10 \mu \text{sec}$.

(B) When the particle enters the magnetic field at $x = a$,it experiences a magnetic force that causes it to move in a circular path of radius $R = \frac{MV}{qB}$.
For the particle to emerge from the region $x = b$,the radius of its circular path must be at least equal to the width of the magnetic field region,which is $(b - a)$.
Therefore,we have the condition: $R \ge (b - a)$.
Substituting the expression for $R$: $\frac{MV}{qB} \ge (b - a)$.
Solving for $V$: $V \ge \frac{q(b - a)B}{M}$.
Thus,the minimum velocity required is $V_{\min} = \frac{q(b - a)B}{M}$.

(B) For the electron to pass without deflection,the electric force must be balanced by the magnetic force: $|\overrightarrow{F}_e| = |\overrightarrow{F}_m|$.
Since $|\overrightarrow{F}_e| = eE$ and $|\overrightarrow{F}_m| = evB$,we have $eE = evB$,which implies $v = \frac{E}{B}$.
The electric field between the plates of a capacitor is given by $E = \frac{\sigma}{\varepsilon_0}$.
Substituting this into the velocity equation,we get $v = \frac{\sigma}{\varepsilon_0 B}$.
The time $t$ taken to travel the length $l$ is $t = \frac{l}{v}$.
Substituting the value of $v$,we get $t = \frac{l}{(\sigma / \varepsilon_0 B)} = \frac{\varepsilon_0 l B}{\sigma}$.

(D) The kinetic energy gained by the electron is given by $\frac{1}{2}mv^2 = qV$,which implies the velocity $v = \sqrt{\frac{2qV}{m}}$.
The magnetic force acting on a charged particle moving in a magnetic field is $F = qvB$.
Substituting the expression for velocity,we get $F = qB\sqrt{\frac{2qV}{m}}$.
From this expression,it is clear that $F \propto \sqrt{V}$.
If the potential is increased to $5V$,the new force $F'$ will be $F' = F \sqrt{\frac{5V}{V}} = \sqrt{5}F$.

(C) For an electron to pass through the region of crossed electric and magnetic fields without deflection,the electric force must be equal to the magnetic force.
$F_e = F_m$
$qE = qvB$
$v = \frac{E}{B}$
Given,potential difference $V = 1000 \ V$ and distance $d = 1 \ cm = 10^{-2} \ m$.
The electric field $E = \frac{V}{d} = \frac{1000}{10^{-2}} = 10^5 \ V/m$.
Given magnetic field $B = 1 \ T$.
Substituting the values,$v = \frac{10^5}{1} = 10^5 \ m/s$.

(D) The force acting on a charged particle moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the Lorentz force formula:
$\vec{F} = q(\vec{v} \times \vec{B})$
The magnitude of this force is given by:
$F = qvB \sin \theta$
where $\theta$ is the angle between the velocity vector $\vec{v}$ and the magnetic field vector $\vec{B}$.
For the force to be non-zero $(F \neq 0)$,we must have $\sin \theta \neq 0$.
Since $\sin \theta = 0$ at $\theta = 0^{\circ}$ and $\theta = 180^{\circ}$,the force is zero in these cases.
Therefore,for the force to be non-zero,the angle $\theta$ can have any value other than $0^{\circ}$ and $180^{\circ}$.

(A) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it experiences a magnetic Lorentz force given by $\vec{F} = Q(\vec{v} \times \vec{B})$.
Since the force is always perpendicular to the velocity of the particle,the magnetic field does no work on the particle $(W = \int \vec{F} \cdot d\vec{r} = 0)$.
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 kinetic energy of the particle remains constant throughout its motion.
Therefore,after $3$ $seconds$,the kinetic energy remains $K$.

(D) The magnetic force $\vec{F}$ on a charged particle is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
Given:
Charge $q = -2 \times 10^{-6}\, C$
Velocity $\vec{v} = (2\hat{i} + 3\hat{j}) \times 10^6\, m/s$
Magnetic field $\vec{B} = 2\hat{j}\, T$
Substituting these values into the formula:
$\vec{F} = (-2 \times 10^{-6}) \times [(2\hat{i} + 3\hat{j}) \times 10^6] \times (2\hat{j})$
$\vec{F} = -2 \times 2 \times [ (2\hat{i} \times \hat{j}) + (3\hat{j} \times \hat{j}) ]$
Since $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{j} = 0$:
$\vec{F} = -4 \times (2\hat{k} + 0)$
$\vec{F} = -8\hat{k}\, N$
The negative sign indicates the force is in the $-z$ direction.
Thus,the force is $8\, N$ in the $-z$ direction.

(C) The net Lorentz force on the particle is given by $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$. For the particle to move in a straight horizontal line,the net force must be zero.
For statement $(2)$: If both $\vec{B}$ and $\vec{E}$ are along the direction of velocity $\vec{v}$,then $\vec{v} \times \vec{B} = 0$. The electric force $q\vec{E}$ acts along the direction of velocity,causing acceleration but no deflection. Thus,the particle continues to move in a straight line.
For statement $(3)$: If $\vec{v}$,$\vec{E}$,and $\vec{B}$ are mutually perpendicular,the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ acts perpendicular to $\vec{v}$. By choosing $\vec{E}$ such that the electric force $\vec{F}_e = q\vec{E}$ is equal and opposite to $\vec{F}_m$,the net force becomes zero. This is the principle of a velocity selector.
Therefore,both $(2)$ and $(3)$ are possible scenarios for maintaining a straight-line path.

(D) The force on an electron due to an electric field is given by $\vec{F}_{E} = -e\vec{E}$.
The force on an electron due to a magnetic field is given by $\vec{F}_{B} = -e(\vec{v} \times \vec{B})$.
Since the velocity $\vec{v}$ and the magnetic field $\vec{B}$ are in the same direction,the angle between them is $0^{\circ}$. Therefore,$\vec{v} \times \vec{B} = 0$,which means $\vec{F}_{B} = 0$.
The total force on the electron is $\vec{F} = \vec{F}_{E} + \vec{F}_{B} = -e\vec{E}$.
Since the electric field $\vec{E}$ acts in the same direction as the velocity,the force $-e\vec{E}$ acts in the opposite direction to the velocity of the electron.
Because the force is opposite to the direction of motion,the speed of the electron will decrease.

(A) The radius $R$ of a circular path of a charged particle with mass $m$,charge $q$,and kinetic energy $K$ in a uniform magnetic field $B$ is given by:
$R = \frac{\sqrt{2mK}}{Bq}$
For a proton: $m_p = m$,$q_p = e$,$K_p = 1\, MeV$.
$R_p = \frac{\sqrt{2m(1)}}{Be}$
For an $\alpha$-particle: $m_{\alpha} = 4m$,$q_{\alpha} = 2e$.
$R_{\alpha} = \frac{\sqrt{2(4m)K_{\alpha}}}{B(2e)} = \frac{\sqrt{8mK_{\alpha}}}{2Be} = \frac{\sqrt{2mK_{\alpha}}}{Be}$
Since the radii are equal $(R_p = R_{\alpha})$:
$\frac{\sqrt{2m(1)}}{Be} = \frac{\sqrt{2mK_{\alpha}}}{Be}$
$1 = K_{\alpha}$
Therefore,the kinetic energy of the $\alpha$-particle is $1\, MeV$.

(A) The radius of a circular orbit for a charged particle in a uniform magnetic field $B$ is given by $R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$, where $K$ is the kinetic energy.
Rearranging for kinetic energy $K$, we get $K = \frac{q^2 B^2 R^2}{2m}$.
For the proton $(p)$ and alpha particle $(\alpha)$, since $B$ and $R$ are equal, the ratio of their kinetic energies is:
$\frac{K_{\alpha}}{K_{p}} = \left(\frac{q_{\alpha}}{q_{p}}\right)^2 \left(\frac{m_{p}}{m_{\alpha}}\right)$.
Given $q_{\alpha} = 2q_{p}$ and $m_{\alpha} = 4m_{p}$, we have:
$\frac{K_{\alpha}}{K_{p}} = (2)^2 \times \left(\frac{1}{4}\right) = 4 \times \frac{1}{4} = 1$.
Therefore, $K_{\alpha} = K_{p} = 1 \, MeV$.

(C) Given: Magnetic field $B = 3.57 \times 10^{-2} \, T$ and specific charge $\frac{e}{m} = 1.76 \times 10^{11} \, C/kg$.
The frequency of revolution $f$ of a charged particle in a magnetic field is given by the formula:
$f = \frac{1}{T} = \frac{qB}{2 \pi m}$
Substituting the given values:
$f = \frac{1}{2 \times 3.14} \times (1.76 \times 10^{11}) \times (3.57 \times 10^{-2})$
$f = \frac{1}{6.28} \times 6.2832 \times 10^9$
$f \approx 1 \times 10^9 \, Hz = 1 \, GHz$.

(C) When a charged particle of mass $m$,charge $q$,and kinetic energy $K$ is projected perpendicular to a uniform magnetic field $B$,it follows a circular path.
The radius $r$ of this 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 $r = \frac{\sqrt{2mK}}{qB}$.
The area $A$ bounded by the circular path is $A = \pi r^2$.
Substituting the expression for $r$,we get $A = \pi \left( \frac{\sqrt{2mK}}{qB} \right)^2 = \frac{\pi (2mK)}{q^2 B^2}$.
Since $m$,$q$,and $B$ are constants,$A \propto K$.

(C) When a charged particle moves in a circle within a magnetic field $B$, it experiences a magnetic force $F_m = qvB$ directed towards the center of the circle.
To make the particle move in a straight line with uniform velocity, the net force on it must be zero.
This is achieved by applying an electric field $E$ such that the electric force $F_e = qE$ exactly balances the magnetic force $F_m$.
Thus, $qE = qvB$, which simplifies to $E = vB$.
The radius of the circular path is given by $r = \frac{mv}{qB}$, so the velocity is $v = \frac{qBr}{m}$.
Substituting this expression for $v$ into the equation for $E$, we get $E = \left(\frac{qBr}{m}\right)B = \frac{qB^2r}{m}$.
Given values: $q = 10.0\,\mu C = 10 \times 10^{-6}\,C$, $m = 1\,\mu g = 1 \times 10^{-9}\,kg$, $r = 10\,cm = 0.1\,m$, and $B = 0.1\,T$.
Substituting these values: $E = \frac{(10 \times 10^{-6}) \times (0.1)^2 \times 0.1}{1 \times 10^{-9}} = \frac{10^{-5} \times 0.01 \times 0.1}{10^{-9}} = \frac{10^{-8}}{10^{-9}} = 10\,V/m$.

(C) The magnetic force on a moving charge is given by the formula $\vec F_m = q(\vec v \times \vec B)$.
Given:
Charge $q = 2.0 \times 10^{-6}\,C$
Velocity $\vec v = 3.0 \times 10^6\,\hat i\,m/s$
Magnetic field $\vec B = -0.2\,\hat k\,T$
Substituting these values into the formula:
$\vec F_m = (2.0 \times 10^{-6}) \times (3.0 \times 10^6\,\hat i \times -0.2\,\hat k)$
$\vec F_m = (2.0 \times 3.0 \times -0.2) \times (\hat i \times \hat k)$
Since $\hat i \times \hat k = -\hat j$,we have:
$\vec F_m = -1.2 \times (-\hat j) = 1.2\,\hat j\,N$
Therefore,the force is $1.2\,N$ along the positive $y$-direction.

(A) $1$. For an electron (negatively charged particle) moving with velocity $v$ in a magnetic field $B$,the magnetic force is given by $F = q(v \times B)$. Since the charge $q$ is negative,the force is in the direction opposite to $(v \times B)$.
$2$. Using Fleming's Left-Hand Rule for a positive charge,the force is perpendicular to both $v$ and $B$. For an electron,we use the Right-Hand Rule or reverse the direction obtained from the Left-Hand Rule.
$3$. In region $B_1$,the electron enters and turns clockwise. By applying the right-hand rule for a negative charge,the magnetic field $B_1$ must be directed into the page.
$4$. In region $B_2$,the electron turns counter-clockwise. By applying the same rule,the magnetic field $B_2$ must be directed out of the page.
$5$. The radius of the circular path is given by $r = \frac{mv}{qB}$. Since $m, v,$ and $q$ are constant,$r \propto \frac{1}{B}$.
$6$. From the figure,the radius of the path in region $B_1$ $(r_1)$ is smaller than the radius of the path in region $B_2$ $(r_2)$.
$7$. Since $r_1 < r_2$,it follows that $B_1 > B_2$. Thus,$B_1$ is stronger than $B_2$.

(B) The net force on the electron is given by the Lorentz force equation: $\vec{F}_{net} = \vec{F}_{e} + \vec{F}_{m} = 0$.
Since $\vec{F}_{e} = q\vec{E}$ and $\vec{F}_{m} = q(\vec{v} \times \vec{B})$,we have $q\vec{E} + q(\vec{v} \times \vec{B}) = 0$,which simplifies to $\vec{E} + (\vec{v} \times \vec{B}) = 0$.
Given $\vec{v} = v\hat{i}$ and $\vec{E} = -E\hat{j}$,substituting these into the equation gives: $-E\hat{j} + (v\hat{i} \times \vec{B}) = 0$,or $v\hat{i} \times \vec{B} = E\hat{j}$.
For the cross product of $\hat{i}$ and $\vec{B}$ to result in $\hat{j}$,the vector $\vec{B}$ must be along the negative $z$-axis $(-\hat{k})$,because $\hat{i} \times (-\hat{k}) = \hat{j}$.
Thus,the magnetic field must be directed along the negative $z$-axis.

(D) The electric field $\vec{E} = E_0 \hat{i}$ exerts a force $\vec{F}_e = qE_0 \hat{i}$,causing acceleration $a_x = \frac{qE_0}{m}$ along the $x$-axis.
The magnetic field $\vec{B} = B_0 \hat{i}$ is parallel to the electric field and the $x$-axis. Since the initial velocity $\vec{v} = v_0 \hat{j}$ is perpendicular to the magnetic field,the particle undergoes helical motion in the $yz$-plane while accelerating along the $x$-axis.
The velocity components at time $t$ are $v_x = a_x t = \frac{qE_0}{m} t$ and the speed in the $yz$-plane remains constant at $v_{\perp} = v_0$.
The total speed $v$ at time $t$ is given by $v = \sqrt{v_x^2 + v_{\perp}^2} = \sqrt{(\frac{qE_0}{m} t)^2 + v_0^2}$.
We are given that the speed becomes $2v_0$ at time $t$:
$2v_0 = \sqrt{(\frac{qE_0}{m} t)^2 + v_0^2}$
Squaring both sides:
$4v_0^2 = (\frac{qE_0}{m} t)^2 + v_0^2$
$3v_0^2 = (\frac{qE_0}{m} t)^2$
Taking the square root:
$\sqrt{3}v_0 = \frac{qE_0}{m} t$
Solving for $t$:
$t = \frac{\sqrt{3}mv_0}{qE_0}$

(B) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path.
The radius of this circular path is given by $R = \frac{mv}{qB_0}$.
The particle is projected from $(d, 0, 0)$ towards the negative $x$-axis. The $y-z$ plane is the plane where $x = 0$.
For the particle not to hit the $y-z$ plane,the radius of the circular path must be less than or equal to the distance $d$ from the $y-z$ plane.
Thus,$R \leq d$.
Substituting the expression for $R$: $\frac{mv}{qB_0} \leq d$.
Solving for $v$: $v \leq \frac{B_0 q d}{m}$.
Therefore,the maximum value of $v$ is $v_{\text{max}} = \frac{B_0 q d}{m}$.

(D) The specific charge is given by $\alpha = q/m$. The particle moves in a helical path in a uniform magnetic field.
The time period of the helical motion is $T = \frac{2 \pi m}{B_o q} = \frac{2 \pi}{B_o \alpha}$.
The given time is $t = \frac{\pi}{B_o \alpha} = \frac{T}{2}$.
The velocity component parallel to the magnetic field is $v_x = V_o$,which remains constant. Thus,the $x$-coordinate at $t = T/2$ is $x = v_x \cdot t = V_o \cdot \frac{\pi}{B_o \alpha} = \frac{V_o \pi}{B_o \alpha}$.
The velocity component perpendicular to the magnetic field is $v_y = V_o$. The particle undergoes circular motion in the $yz$-plane with radius $r = \frac{m v_y}{B_o q} = \frac{V_o}{B_o \alpha}$.
At $t = T/2$,the particle completes half a circle in the $yz$-plane. Starting from $(0, 0)$ in the $yz$-plane with initial velocity in the $+y$ direction,after half a period,the displacement in the $y$-direction is $0$ and in the $z$-direction is $-2r = -\frac{2 V_o}{B_o \alpha}$.
Therefore,the coordinates are $\left( \frac{V_o \pi}{B_o \alpha}, 0, - \frac{2 V_o}{B_o \alpha} \right)$.

(B) The radius of the circular path of a charged particle in a uniform 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: $r = \frac{m}{qB} \sqrt{\frac{2K}{m}} = \frac{\sqrt{2mK}}{qB}$.
For constant $K$ and $B$,the radius $r \propto \frac{\sqrt{m}}{q}$.
Deflection is inversely proportional to the radius $(d \propto \frac{1}{r})$,so $d \propto \frac{q}{\sqrt{m}}$.
For $H^+$: $q=1, m=1 \Rightarrow d \propto \frac{1}{\sqrt{1}} = 1$.
For $He^+$: $q=1, m=4 \Rightarrow d \propto \frac{1}{\sqrt{4}} = 0.5$.
For $O^{2+}$: $q=2, m=16 \Rightarrow d \propto \frac{2}{\sqrt{16}} = \frac{2}{4} = 0.5$.
Comparing the values,$H^+$ has the highest deflection,while $He^+$ and $O^{2+}$ have equal and lower deflection.

(C) The radius $r$ of a charged particle moving in a uniform magnetic field is given by $r = \frac{mv}{qB}$.
Since kinetic energy $T = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mT}$.
Substituting this into the radius formula,we get $r = \frac{\sqrt{2mT}}{qB}$.
Given the initial state: $R = \frac{\sqrt{2mT}}{qB}$.
For the new state,kinetic energy $T' = 2T$ and magnetic field $B' = 3B$.
The new radius $R'$ is given by $R' = \frac{\sqrt{2m(2T)}}{q(3B)}$.
$R' = \frac{\sqrt{2} \cdot \sqrt{2mT}}{3qB} = \frac{\sqrt{2}}{3} \left( \frac{\sqrt{2mT}}{qB} \right)$.
Substituting $R$ for the term in the parentheses,we get $R' = \frac{\sqrt{2}}{3} R = \sqrt{\frac{2}{9}} R$.