Motion of Charged Particle In Magnetic Field Questions in English

(C) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Since the magnetic field $\vec{B}$ is perpendicular to the plane of the orbit,the force $\vec{F}$ acts towards the center of the circular path (centripetal force).
The torque $\vec{\tau}$ about the nucleus is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
Since the force $\vec{F}$ is directed towards the center,the position vector $\vec{r}$ and the force vector $\vec{F}$ are collinear (anti-parallel).
Therefore,$\vec{\tau} = \vec{r} \times \vec{F} = 0$.
Since the torque is zero,the dot product of the torque vector with any vector,including the angular momentum vector $\vec{L}$,must be zero: $\vec{\tau} \cdot \vec{L} = 0$.

(A, C) The particle enters region-$c$ if the radius of its circular path $R$ is greater than the width $L$ of region-$b$.
Since $R = \frac{mv}{qB}$,the condition for entering region-$c$ is $\frac{mv}{qB} > L$,which implies $v > \frac{qBL}{m}$. Thus,option $A$ is correct.
Since the magnetic field $B$ is uniform and the velocity vector $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,the magnetic force provides the necessary centripetal force,making the path of the particle circular in region-$b$. Thus,option $C$ is correct.
The time spent in region-$b$ is given by $t = \frac{\theta}{\omega}$,where $\theta$ is the angle subtended by the arc at the center and $\omega = \frac{qB}{m}$ is the angular velocity. Since the angle $\theta$ depends on the radius $R$ (and thus on velocity $v$),the time spent in region-$b$ depends on the velocity $v$. Therefore,option $D$ is incorrect.

(D) When a charged particle moves with velocity $v$ perpendicular to a uniform magnetic field $B$ (i.e.,$\theta = 90^{\circ}$),it experiences a magnetic Lorentz force $F = qvB$.
This force acts perpendicular to the velocity,providing the necessary centripetal force for circular motion:
$qvB = \frac{mv^2}{r}$
Solving for the radius $r$:
$r = \frac{mv}{qB}$
Since the momentum of the particle is $p = mv$,we can write:
$r = \frac{p}{qB}$
From this expression,it is clear that the radius $r$ is directly proportional to the momentum $p$ of the particle.
Therefore,option $(D)$ is correct.

(A) According to the question,the charge on the particle is $q$ and its velocity is $v$.
Due to the uniform electric field,the electric force on the particle is $F_{\text{electric}} = qE$.
Due to the uniform magnetic field,the magnetic force on the particle is given by $F_{\text{magnetic}} = q(v \times B)$.
For the particle to move undeflected,the net Lorentz force must be zero,which means the electric force and magnetic force must be equal in magnitude and opposite in direction: $F_{\text{electric}} = F_{\text{magnetic}}$.
Substituting the magnitudes,we get $qE = qvB$.
Solving for velocity,we find $v = \frac{E}{B}$.
Thus,the condition for the particle to move undeflected is $v = \frac{E}{B}$.

(B) The time period $T$ of a charged particle moving in a circular path in a uniform magnetic field is given by the formula: $T = \frac{2\pi m}{qB}$.
This formula shows that the time period $T$ depends only on the mass $m$,the charge $q$,and the magnetic field strength $B$.
Since $m$,$q$,and $B$ remain constant,the time period $T$ is independent of the speed $v$ of the proton.
Therefore,even if the speed is increased to $\sqrt{2}v$,the time period will remain $T$.

(A) Given:
Velocity of proton $v = 10^{6} \ m/s$ along the $Y$-axis.
Electric field $E = 2 \times 10^{4} \ V/m$ along the negative $X$-axis.
Specific charge $\frac{e}{m} = 10^{8} \ C/kg$.
Since the proton moves with a uniform velocity,the net force on it is zero. Thus,the magnetic force balances the electric force:
$qE = qvB$
$B = \frac{E}{v} = \frac{2 \times 10^{4}}{10^{6}} = 2 \times 10^{-2} \ T$
When the electric field is switched off,the proton moves in a circular path due to the magnetic force acting as the centripetal force:
$R = \frac{mv}{qB} = \frac{v}{(e/m)B}$
$R = \frac{10^{6}}{10^{8} \times 2 \times 10^{-2}}$
$R = \frac{10^{6}}{2 \times 10^{6}} = 0.5 \ m$
Therefore,the radius of the circle is $0.5 \ m$.

(B) The radius $R$ of the circular path followed by a charged particle in a magnetic field is given by $R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$,where $K$ is the kinetic energy.
Since the particle is accelerated through a potential difference $V$,its kinetic energy is $K = qV$.
Substituting this into the radius formula,we get $R = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.
Given that $q$,$V$,and $B$ are the same for both particles,we have $R \propto \sqrt{m}$.
Therefore,$\frac{R_{1}}{R_{2}} = \sqrt{\frac{m_{A}}{m_{B}}}$.
Squaring both sides,we get $\frac{m_{A}}{m_{B}} = \left(\frac{R_{1}}{R_{2}}\right)^{2}$.

(B) Given: Kinetic energy $= E$,Mass $= m$,Magnetic field $= B$,Charge $= q$.
When a charged particle moves perpendicular to a uniform magnetic field,the magnetic force provides the necessary centripetal force for circular motion.
$F_m = qvB$ and $F_c = \frac{mv^2}{r}$.
Equating these,we get $qvB = \frac{mv^2}{r}$,which simplifies to $r = \frac{mv}{qB}$.
We know that kinetic energy $E = \frac{1}{2}mv^2$,so $v = \sqrt{\frac{2E}{m}}$.
Substituting the value of $v$ into the radius formula:
$r = \frac{m}{qB} \sqrt{\frac{2E}{m}} = \frac{\sqrt{m^2} \cdot \sqrt{2E}}{\sqrt{m} \cdot qB} = \frac{\sqrt{2Em}}{qB}$.

(A) For a charged particle moving with velocity $v$ in a region with electric field $E$ and magnetic field $B$,the Lorentz force is $F = q(E + v \times B)$.
For the particle to pass undeflected,the net force must be zero,i.e.,$qE = -q(v \times B)$.
For a proton $(q > 0)$: The electric force $F_E = qE$ is downwards. The magnetic force $F_B = q(v \times B)$ is upwards (using the right-hand rule,$v$ is right,$B$ is out,so $v \times B$ is up). For no deflection,$qE = qvB$,which gives $v = E/B$.
For an electron $(q < 0)$: The electric force $F_E = qE$ is upwards. The magnetic force $F_B = q(v \times B)$ is downwards (since $q$ is negative,the direction of $F_B$ reverses). For no deflection,$|q|E = |q|vB$,which also gives $v = E/B$.
However,the question asks for the behavior of the stream. Since both require the same speed $v = E/B$ to pass undeflected,option $B$ is incorrect as it implies electrons would not pass. Option $A$ is the correct statement.

(B) The magnetic Lorentz force experienced by a charged particle moving in a magnetic field is $F = qvB$.
This magnetic force provides the necessary centripetal force for circular motion: $qvB = \frac{mv^2}{R}$.
Solving for the radius $R$,we get $R = \frac{mv}{qB}$.
The period of revolution $T$ is defined as the time taken to complete one full circle,given by $T = \frac{2\pi R}{v}$.
Substituting the expression for $R$ into the formula for $T$,we get $T = \frac{2\pi}{v} \cdot \frac{mv}{qB} = \frac{2\pi m}{qB}$.
Since $T$ depends only on the mass $m$,charge $q$,and magnetic field $B$,it is independent of both the linear speed $v$ and the radius $R$.

(B) The particle undergoes helical motion in a uniform magnetic field.
The velocity component parallel to the magnetic field is $v_\parallel = 2 \times 10^{-2} \text{ m/s}$ (along $\hat{k}$).
The time period of one revolution is $T = \frac{2\pi m}{qB}$.
Given: $m = 5 \text{ mg} = 5 \times 10^{-6} \text{ kg}$,$q = 5 \times 10^{-6} \text{ C}$,$B = 0.1 \text{ T}$.
$T = \frac{2 \times \pi \times 5 \times 10^{-6}}{5 \times 10^{-6} \times 0.1} = \frac{2\pi}{0.1} = 20\pi \text{ s}$.
The pitch of the helix is $p = v_\parallel \times T = (2 \times 10^{-2}) \times (20\pi) = 0.4\pi \text{ m}$.
For $5$ revolutions,the total distance $\alpha$ along the $\hat{k}$ direction is $\alpha = 5 \times p = 5 \times 0.4\pi = 2\pi \text{ m}$.
Using $\pi \approx 3.14$,$\alpha = 2 \times 3.14 = 6.28 \text{ m}$.

(B) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
First,calculate the cross product $\vec{v} \times \vec{B}$:
$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix}$
$= \hat{i}((-2)(-5) - (3)(3)) - \hat{j}((1)(-5) - (3)(2)) + \hat{k}((1)(3) - (-2)(2))$
$= \hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4)$
$= 1\hat{i} + 11\hat{j} + 7\hat{k}$.
The magnitude of the cross product is $|\vec{v} \times \vec{B}| = \sqrt{1^2 + 11^2 + 7^2} = \sqrt{1 + 121 + 49} = \sqrt{171}$.
Given $q = 1 \mu\text{C} = 10^{-6} \text{ C}$,the force magnitude is $F = |q| |\vec{v} \times \vec{B}| = 10^{-6} \times \sqrt{171} \text{ N}$.
Comparing this with $\sqrt{\alpha} \times 10^{-6} \text{ N}$,we get $\alpha = 171$.

(B) The magnetic force acting on a charged particle is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
According to Newton's second law,$\vec{F} = m\vec{a}$,so $m\vec{a} = q(\vec{v} \times \vec{B})$.
Since the magnetic force $\vec{F}$ is always perpendicular to the magnetic field $\vec{B}$,the acceleration $\vec{a}$ must also be perpendicular to the magnetic field $\vec{B}$.
Therefore,the dot product of acceleration and magnetic field must be zero: $\vec{a} \cdot \vec{B} = 0$.
Given $\vec{B} = (3\hat{i} + 2\hat{j}) \ \text{T}$ and $\vec{a} = (4\hat{i} - \frac{x}{2}\hat{j}) \ \text{m/s}^2$,we calculate the dot product:
$(4)(3) + (-\frac{x}{2})(2) = 0$
$12 - x = 0$
$x = 12$.