Question

The ratio of their kinetic energies $\left(\frac{K_1}{K_2}\right)$ is :

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $4$
• D. $1$
• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $\frac{1}{2}$
• D. $2$
• A. both particles revolve clockwise
• B. both particles revolve anticlockwise
• C. particle 1 revolves clockwise while particle 2 revolves anticlockwise
• D. particle 1 revolves anticlockwise while particle 2 revolves clockwise
• A. 1 s
• B. 2 s
• C. 4 s
• D. 8 s
• A. $p_1 = \frac{p_2}{2}$
• B. $p_1 = p_2$
• C. $p_1 = 2p_2$
• D. $p_1 = 4p_2$

Answer 1

The Correct Option is A

Concept:
• When a particle with charge $q$ is accelerated from rest across a potential difference $V$, the work done entirely transforms into the particle's kinetic energy.
• The kinetic energy $K$ acquired is governed strictly by the formula: $K = |q|V$, where $|q|$ is the magnitude of the charge.
• This relationship is notably completely independent of the mass of the accelerated particle.

Step 1: Calculate kinetic energy for Particle 1
Particle 1 has a charge of $-q$. Since it is being accelerated, we utilize the magnitude of the charge to calculate the positive kinetic energy gained.
The accelerating potential difference is $V$.
\[ K_1 = |-q| \times V = qV \]

Step 2: Calculate kinetic energy for Particle 2
Particle 2 has a charge of $+2q$.
The accelerating potential difference is exactly the same, $V$.
\[ K_2 = |+2q| \times V = 2qV \]

Step 3: Determine the ratio
We are asked to find the specific ratio $\frac{K_1}{K_2}$.
\[ \frac{K_1}{K_2} = \frac{qV}{2qV} \]
Cancel the common terms $q$ and $V$ from the numerator and denominator:
\[ \frac{K_1}{K_2} = \frac{1}{2} \]

Step 4: Conclusion
The ratio of their acquired kinetic energies is strictly $1/2$. This elegantly corresponds to option (A).

Answer 2

Concept:
• When a charged particle enters a uniform magnetic field perpendicularly, it traces a circular trajectory.
• The radius $r$ of this circular path is heavily dependent on the particle's momentum $p$, charge $q$, and the magnetic field $B$: $r = \frac{p}{qB}$.
• Momentum can be smoothly linked back to kinetic energy $K$ via the relation $p = \sqrt{2mK}$.
• Combining these provides a comprehensive radius formula: $r = \frac{\sqrt{2mK}}{qB}$.

Step 1: Derive expression for the radius of Particle 1
For Particle 1:
Mass = $m$
Charge magnitude = $q$
Kinetic energy = $K_1 = qV$
Apply the comprehensive radius formula:
\[ r_1 = \frac{\sqrt{2 m K_1}}{q B} \]
\[ r_1 = \frac{\sqrt{2 m (qV)}}{q B} = \frac{\sqrt{2 m q V}}{q B} \]

Step 2: Derive expression for the radius of Particle 2
For Particle 2:
Mass = $\frac{m}{2}$
Charge magnitude = $2q$
Kinetic energy = $K_2 = 2qV$
Apply the comprehensive radius formula:
\[ r_2 = \frac{\sqrt{2 \left(\frac{m}{2}\right) K_2}}{(2q) B} \]
\[ r_2 = \frac{\sqrt{2 \left(\frac{m}{2}\right) (2qV)}}{2q B} \]
\[ r_2 = \frac{\sqrt{m (2qV)}}{2q B} = \frac{\sqrt{2 m q V}}{2q B} \]

Step 3: Calculate the ratio of the radii
Now, divide the expression for $r_1$ by the expression for $r_2$:
\[ \frac{r_1}{r_2} = \frac{ \left( \frac{\sqrt{2 m q V}}{q B} \right) }{ \left( \frac{\sqrt{2 m q V}}{2q B} \right) } \]
The massive square root terms ($\sqrt{2 m q V}$) and the $B$ terms cancel out entirely:
\[ \frac{r_1}{r_2} = \frac{ \left( \frac{1}{q} \right) }{ \left( \frac{1}{2q} \right) } \]
\[ \frac{r_1}{r_2} = \frac{1}{q} \times \frac{2q}{1} = 2 \]

Step 4: Conclusion
The ratio of their circular radii $\left(\frac{r_1}{r_2}\right)$ is exactly 2. This perfectly matches option (D).

Answer 3

Concept:
• The direction of the magnetic force $\vec{F}$ acting on a moving charged particle is strictly determined by the Lorentz force vector cross product: $\vec{F} = q(\vec{v} \times \vec{B})$.
• The resultant direction of this force dictates the initial bending direction of the particle's path, and consequently, its direction of circular revolution (clockwise or anticlockwise).
• We evaluate this from the standard perspective, typically looking down directly from the $+z$ axis towards the xy-plane.

Step 1: Determine force and direction for Particle 1
Particle 1 has a strictly negative charge: $q_1 = -q$.
Its initial velocity vector is entirely along the +x axis: $\vec{v}_1 = v_1 \hat{i}$.
The uniform magnetic field vector is entirely along the +z axis: $\vec{B} = B_0 \hat{k}$.
Apply the Lorentz force equation:
\[ \vec{F}_1 = (-q) (\vec{v}_1 \times \vec{B}) \]
\[ \vec{F}_1 = (-q) [(v_1 \hat{i}) \times (B_0 \hat{k})] \]
Using the right-hand rule for unit vectors, $\hat{i} \times \hat{k} = -\hat{j}$.
\[ \vec{F}_1 = -q \cdot v_1 B_0 (-\hat{j}) \]
\[ \vec{F}_1 = +q v_1 B_0 \hat{j} \]
The force points strongly in the $+y$ direction. An initial velocity in the $+x$ direction being violently pulled toward the $+y$ direction causes the particle to curve leftwards. Looking from the $+z$ axis, this curving path forms an anticlockwise (counter-clockwise) revolution in the xy-plane.

Step 2: Determine force and direction for Particle 2
Particle 2 has a strictly positive charge: $q_2 = +2q$.
Its initial velocity vector is identical in direction: $\vec{v}_2 = v_2 \hat{i}$.
The uniform magnetic field remains: $\vec{B} = B_0 \hat{k}$.
Apply the Lorentz force equation:
\[ \vec{F}_2 = (+2q) (\vec{v}_2 \times \vec{B}) \]
\[ \vec{F}_2 = (+2q) [(v_2 \hat{i}) \times (B_0 \hat{k})] \]
Again, $\hat{i} \times \hat{k} = -\hat{j}$.
\[ \vec{F}_2 = +2q \cdot v_2 B_0 (-\hat{j}) \]
\[ \vec{F}_2 = -2q v_2 B_0 \hat{j} \]
The force points strongly in the $-y$ direction. An initial velocity in the $+x$ direction being violently pulled toward the $-y$ direction causes the particle to curve rightwards. Looking from the $+z$ axis, this curving path forms a clockwise revolution in the xy-plane.

Step 3: Conclusion
Particle 1 revolves securely in an anticlockwise direction, and particle 2 revolves securely in a clockwise direction. This precisely matches option (D).

Answer 4

Concept:
• The time period ($T$) of a charged particle revolving in a uniform magnetic field is the time taken to complete one full circular orbit.
• It depends exclusively on the particle's mass $m$, charge magnitude $q$, and the magnetic field $B$.
• The standard derived formula is $T = \frac{2\pi m}{qB}$.
• Notably, the time period is entirely independent of the particle's velocity or kinetic energy.

Step 1: Establish the time period for Particle 1
For Particle 1, mass is $m$ and charge magnitude is $q$.
The given time period is $T_1 = 4 \text{ s}$.
According to the standard formula:
\[ T_1 = \frac{2\pi m}{q B} = 4 \text{ s} \]

Step 2: Derive the time period formula for Particle 2
For Particle 2, the mass is given as $\frac{m}{2}$ and the charge magnitude is $2q$.
Apply these specific values into the standard time period formula to find $T_2$:
\[ T_2 = \frac{2\pi \left(\frac{m}{2}\right)}{(2q) B} \]
Simplify the complex fraction:
\[ T_2 = \frac{\pi m}{2q B} \]

Step 3: Relate $T_2$ to $T_1$ and calculate
We can rewrite $T_2$ to expose the structure of $T_1$ within it:
\[ T_2 = \frac{1}{4} \times \left( \frac{2\pi m}{q B} \right) \]
Since we know that the parenthetical term is exactly $T_1$:
\[ T_2 = \frac{1}{4} \times T_1 \]
Substitute the known value of $T_1 = 4 \text{ s}$:
\[ T_2 = \frac{1}{4} \times 4 \text{ s} \]
\[ T_2 = 1 \text{ s} \]

Step 4: Conclusion
The period of revolution for particle 2 will be firmly established as $1 \text{ s}$. This matches option (A).

Answer 5

Concept:
• When a charged particle is accelerated from rest by an electric potential difference $V$, it acquires kinetic energy $K = qV$.
• This acquired kinetic energy can be smoothly translated into the particle's momentum $p$.
• The foundational relationship between momentum, mass, and kinetic energy is $p = \sqrt{2mK}$.

Step 1: Calculate momentum for Particle 1
Particle 1 has mass $m$ and acquired kinetic energy $K_1 = qV$.
Utilize the momentum-energy relation:
\[ p_1 = \sqrt{2 \cdot m \cdot K_1} \]
Substitute the kinetic energy:
\[ p_1 = \sqrt{2 m (qV)} = \sqrt{2mqV} \]

Step 2: Calculate momentum for Particle 2
Particle 2 has a distinctly different mass $\frac{m}{2}$ and acquired kinetic energy $K_2 = 2qV$.
Utilize the momentum-energy relation:
\[ p_2 = \sqrt{2 \cdot \left(\frac{m}{2}\right) \cdot K_2} \]
Substitute the kinetic energy:
\[ p_2 = \sqrt{2 \cdot \left(\frac{m}{2}\right) \cdot (2qV)} \]
Simplify the terms inside the massive square root:
The $2$ and the $\frac{1}{2}$ conveniently cancel each other out, leaving:
\[ p_2 = \sqrt{m \cdot (2qV)} = \sqrt{2mqV} \]

Step 3: Compare the two momentums
Comparing the completely simplified expressions from Step 1 and
Step 2:
$p_1 = \sqrt{2mqV}$
$p_2 = \sqrt{2mqV}$
It is mathematically obvious that $p_1$ is exactly equal to $p_2$.

Step 4: Conclusion
Despite having vastly different masses and charges, their resulting momentums are entirely identical due to how the mass and charge ratios perfectly canceled out. Thus, $p_1 = p_2$, corresponding to option (B).