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. \( \frac{1}{2} \)
• B. \( \sqrt{2} \)
• C. \( \frac{1}{\sqrt{2}} \)
• 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. \( 1\,\text{s} \)
• B. \( 2\,\text{s} \)
• C. \( 4\,\text{s} \)
• D. \( 8\,\text{s} \)
• 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 carrying a net electric charge \(q\) is accelerated from rest through an electric potential difference \(V\), the work done on the particle by the electrostatic field is given by \(W = qV\). According to the Work-Energy Theorem, this work done is converted entirely into the particle's kinetic energy (\(K\)): \[ K = q \cdot V \] Where \(q\) is the magnitude of the charge of the particle, and \(V\) is the accelerating potential difference.

Step 1: Formulate the kinetic energy expression for the first particle. Particle 1 has a charge magnitude of \( |-q| = q \) and is accelerated through a potential difference \( V \): \[ K_1 = q \cdot V \quad \cdots (1) \]

Step 2: Formulate the kinetic energy expression for the second particle. Particle 2 has a charge magnitude of \( |+2q| = 2q \) and is accelerated through the identical potential difference \( V \): \[ K_2 = (2q) \cdot V = 2qV \quad \cdots (2) \]

Step 3: Set up the ratio of the two kinetic energies by dividing equation (1) by equation (2): \[ \frac{K_1}{K_2} = \frac{q \cdot V}{2q \cdot V} \]

Step 4: Cancel out the common terms \( q \) and \( V \) from both the numerator and the denominator: \[ \frac{K_1}{K_2} = \frac{1}{2} \] Thus, the ratio of their kinetic energies is precisely \( \frac{1}{2} \), which matches Option (A).

Answer 2

Concept: When a charged particle enters a uniform magnetic field \(\vec{B}\) with a velocity \(\vec{v}\) perpendicular to the field lines, it experiences a magnetic Lorentz force (\(F_m = qvB\)) that acts as a centripetal force. This forces the particle into a stable circular trajectory. Equating the forces: \[ \frac{m v^2}{r} = q v B \quad \Rightarrow \quad r = \frac{m v}{q B} = \frac{p}{q B} \] Since the linear momentum \(p\) can be written in terms of kinetic energy \(K\) as \(p = \sqrt{2mK}\), the formula for the radius is: \[ r = \frac{\sqrt{2mK}}{qB} \]

Step 1: Set up the radius expression for Particle 1. Substituting its parameters (\(m_1 = m\), \(K_1 = qV\), charge magnitude \(= q\)): \[ r_1 = \frac{\sqrt{2m(qV)}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}} \quad \cdots (1) \]

Step 2: Set up the radius expression for Particle 2. Substituting its parameters (\(m_2 = \frac{m}{2}\), \(K_2 = 2qV\), charge magnitude \(= 2q\)): \[ r_2 = \frac{\sqrt{2\left(\frac{m}{2}\right)(2qV)}}{2qB} = \frac{\sqrt{2mqV}}{2qB} = \frac{1}{2B}\sqrt{\frac{2mqV}{q^2}} = \frac{1}{2B}\sqrt{\frac{2mV}{q}} \quad \cdots (2) \]

Step 3: Divide equation (1) by equation (2) to find the ratio of their trajectories: \[ \frac{r_1}{r_2} = \frac{\frac{1}{B}\sqrt{\frac{2mV}{q}}}{\frac{1}{2B}\sqrt{\frac{2mV}{q}}} \]

Step 4: Cancel out the common terms \(\frac{1}{B}\sqrt{\frac{2mV}{q}}\) from the expression: \[ \frac{r_1}{r_2} = \frac{1}{\left(\frac{1}{2}\right)} = 2 \] The ratio of the radii of their circular paths is exactly \( 2 \), which matches Option (D).

Answer 3

Concept: The directional vector of the magnetic force acting upon a moving charge is determined by the vector cross product rule defined by the Lorentz Force equation: \[ \vec{F} = q \cdot (\vec{v} \times \vec{B}) \] Where \(q\) includes the algebraic sign of the particle's charge. The cross product of the unit vectors along the Cartesian axes gives: \[ \hat{i} \times \hat{k} = -\hat{j} \]

Step 1: Determine the direction of the magnetic force acting on Particle 1. Particle 1 has a negative charge (\(q_1 = -q\)), initial velocity vector \(\vec{v}_1 = v_1 \hat{i}\), and magnetic field vector \(\vec{B} = B_0 \hat{k}\): \[ \vec{F}_1 = (-q) \cdot (v_1 \hat{i} \times B_0 \hat{k}) = (-q \cdot v_1 \cdot B_0) \cdot (\hat{i} \times \hat{k}) \] Substituting \(\hat{i} \times \hat{k} = -\hat{j}\): \[ \vec{F}_1 = (-q \cdot v_1 \cdot B_0) \cdot (-\hat{j}) = +q v_1 B_0 \hat{j} \] Since the particle moves in the \(+\hat{i}\) direction and experiences a deflecting force toward \(+\hat{j}\), it curves upward in the \(xy\)-plane, which corresponds to an anticlockwise rotation.

Step 2: Determine the direction of the magnetic force acting on Particle 2. Particle 2 has a positive charge (\(q_2 = +2q\)), initial velocity vector \(\vec{v}_2 = v_2 \hat{i}\), and magnetic field vector \(\vec{B} = B_0 \hat{k}\): \[ \vec{F}_2 = (+2q) \cdot (v_2 \hat{i} \times B_0 \hat{k}) = (2q \cdot v_2 \cdot B_0) \cdot (\hat{i} \times \hat{k}) \] Substituting \(\hat{i} \times \hat{k} = -\hat{j}\): \[ \vec{F}_2 = (2q \cdot v_2 \cdot B_0) \cdot (-\hat{j}) = -2q v_2 B_0 \hat{j} \] Since this particle moves in the \(+\hat{i}\) direction and experiences a deflecting force toward \(-\hat{j}\), it curves downward in the \(xy\)-plane, which corresponds to a clockwise rotation.

Step 3: Combine the two directional results. Particle 1 revolves anticlockwise, while particle 2 revolves clockwise, matching Option (D).

Answer 4

Concept: The time period of revolution (\(T\)) of a charged particle moving along a circular path in a uniform magnetic field is the time taken to complete one full orbit. It is equal to the total circumference divided by the orbital speed: \[ T = \frac{2\pi r}{v} = \frac{2\pi \left(\frac{mv}{qB}\right)}{v} = \frac{2\pi m}{qB} \] This reveals that the time period of revolution depends exclusively on the mass-to-charge ratio (\(\frac{m}{q}\)) of the particle and the field strength, and is entirely independent of its velocity or orbital radius.

Step 1: Write out the given physical parameters for both particles:
• Particle 1: \(\text{mass} = m\), \(\text{charge magnitude} = q\), \(\text{Time period } T_1 = 4\,\text{s}\)
• Particle 2: \(\text{mass} = \frac{m}{2}\), \(\text{charge magnitude} = 2q\), \(\text{Time period } = T_2\)

Step 2: Construct the explicit ratio of the two time periods using the formula: \[ \frac{T_2}{T_1} = \frac{\frac{2\pi m_2}{q_2 B}}{\frac{2\pi m_1}{q_1 B}} = \frac{m_2}{m_1} \cdot \frac{q_1}{q_2} \]

Step 3: Substitute the values of masses and charges into this ratio equation: \[ \frac{T_2}{T_1} = \frac{\left(\frac{m}{2}\right)}{m} \cdot \frac{q}{2q} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \]

Step 4: Calculate the value of \(T_2\) using the given value \(T_1 = 4\,\text{s}\): \[ T_2 = \frac{1}{4} \cdot T_1 = \frac{1}{4} \times 4\,\text{s} = 1\,\text{s} \] The time period for particle 2 is exactly \( 1\,\text{s} \), which corresponds to Option (A).

Answer 5

Concept: The linear momentum \(p\) of any particle is linked directly to its mass \(m\) and its kinetic energy \(K\) through the classical mechanics formula: \[ p = \sqrt{2mK} \] By using the expression for kinetic energy acquired from a potential difference (\(K = qV\)), the momentum can be rewritten as: \[ p = \sqrt{2mqV} \]

Step 1: Determine the momentum expression for Particle 1. Substitute \(m_1 = m\) and \(q_1 = q\): \[ p_1 = \sqrt{2 \cdot m \cdot q \cdot V} = \sqrt{2mqV} \quad \cdots (1) \]

Step 2: Determine the momentum expression for Particle 2. Substitute \(m_2 = \frac{m}{2}\) and \(q_2 = 2q\): \[ p_2 = \sqrt{2 \cdot m_2 \cdot q_2 \cdot V} = \sqrt{2 \cdot \left(\frac{m}{2}\right) \cdot (2q) \cdot V} \]

Step 3: Simplify inside the radical for Particle 2. The factor of 2 in the numerator cancels out the factor of 2 in the denominator: \[ p_2 = \sqrt{m \cdot 2q \cdot V} = \sqrt{2mqV} \quad \cdots (2) \]

Step 4: Compare the two expressions. Comparing equation (1) and equation (2) shows that both quantities are identical: \[ p_1 = p_2 \] Therefore, the two particles carry equal linear momentum, corresponding precisely to Option (B).