Question:
In a vacuum chamber, a particle of charge \(1\,\mu\text{C}\) and mass \(1\,\text{mg}\) is projected with a velocity \((\hat{i}+2\hat{j})\,\text{ms}^{-1}\) from the \(XZ\)-plane at time \(t=0\) in an electric field of \(1\hat{i}\,\text{Vm}^{-1}\). At \(t=0.2\,\text{s}\), the electric field is switched off and a magnetic field of \(6\hat{j}\,\text{T}\) is switched on. The acceleration due to gravity is \(-10\hat{j}\,\text{ms}^{-2}\). Correct option(s) is/are:
In a vacuum chamber, a particle of charge \(1\,\mu\text{C}\) and mass \(1\,\text{mg}\) is projected with a velocity \((\hat{i}+2\hat{j})\,\text{ms}^{-1}\) from the \(XZ\)-plane at time \(t=0\) in an electric field of \(1\hat{i}\,\text{Vm}^{-1}\). At \(t=0.2\,\text{s}\), the electric field is switched off and a magnetic field of \(6\hat{j}\,\text{T}\) is switched on. The acceleration due to gravity is \(-10\hat{j}\,\text{ms}^{-2}\). Correct option(s) is/are:
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In combined field problems, check which forces are conservative and which are non-conservative. Magnetic forces do no work, so they only change the direction of velocity, simplifying energy calculations.
Updated On: May 20, 2026
- The vertical distance of the particle from the $XZ$ plane at $t = 0.3 \text{ s}$ is $15 \text{ cm}$.
- The vertical distance of the particle from the $XZ$ plane at $t = 0.4 \text{ s}$ is $10 \text{ cm}$.
- The radius of the trajectory of the particle for $t > 0.2 \text{ s}$ is $20 \text{ cm}$.
- The particle will be in the $XZ$ plane at $t = 0.35 \text{ s}$.
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The particle undergoes two phases of motion: accelerated motion due to $E$ and gravity for $t < 0.2\text{ s}$, and motion under $B$ and gravity for $t > 0.2\text{ s}$. We need to find the state of the particle at $t = 0.2\text{ s}$ to determine its subsequent trajectory.
Step 2: Key Formula or Approach:
• $t < 0.2$: $\vec{a} = \frac{q\vec{E}}{m} + \vec{g}$.
• $t > 0.2$: Magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$. Circular motion radius $R = \frac{mv_{\perp}}{qB}$.
Step 3: Detailed Explanation:
• Motion for $0 \le t \le 0.2$: $q/m = 10^{-6} / 10^{-6} = 1 \text{ C/kg}$. $\vec{a} = (1\hat{i}) + (-10\hat{j}) \text{ m/s}^2$. $\vec{v}(0.2) = (\hat{i} + 2\hat{j}) + (1\hat{i} - 10\hat{j})(0.2) = 1.2\hat{i} + 0\hat{j} \text{ m/s}$. Position $y(0.2) = u_y t + \frac{1}{2} a_y t^2 = 2(0.2) - 5(0.2)^2 = 0.4 - 0.2 = 0.2 \text{ m}$.
• Motion for $t > 0.2$: Magnetic field $\vec{B} = 6\hat{j}$. Since gravity is also in $\hat{j}$, the vertical motion ($y$) is independent of the magnetic force ($\vec{v} \times \vec{B}$ has no $\hat{j}$ component). $y(t) = y(0.2) + v_y(0.2)(t-0.2) - 5(t-0.2)^2 = 0.2 - 5(t-0.2)^2$. At $t = 0.3$, $y = 0.2 - 5(0.1)^2 = 0.15 \text{ m} = 15 \text{ cm}$. (A) is correct. At $t = 0.4$, $y = 0.2 - 5(0.2)^2 = 0 \text{ cm}$. (B) is incorrect.
• Radius of Trajectory: The velocity $\vec{v}(0.2) = 1.2\hat{i}$ is perpendicular to $\vec{B} = 6\hat{j}$. $R = \frac{mv}{qB} = \frac{1.2}{1 \times 6} = 0.2 \text{ m} = 20 \text{ cm}$. (C) is correct.
Step 4: Final Answer:
The correct options are (A) and (C).
The particle undergoes two phases of motion: accelerated motion due to $E$ and gravity for $t < 0.2\text{ s}$, and motion under $B$ and gravity for $t > 0.2\text{ s}$. We need to find the state of the particle at $t = 0.2\text{ s}$ to determine its subsequent trajectory.
Step 2: Key Formula or Approach:
• $t < 0.2$: $\vec{a} = \frac{q\vec{E}}{m} + \vec{g}$.
• $t > 0.2$: Magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$. Circular motion radius $R = \frac{mv_{\perp}}{qB}$.
Step 3: Detailed Explanation:
• Motion for $0 \le t \le 0.2$: $q/m = 10^{-6} / 10^{-6} = 1 \text{ C/kg}$. $\vec{a} = (1\hat{i}) + (-10\hat{j}) \text{ m/s}^2$. $\vec{v}(0.2) = (\hat{i} + 2\hat{j}) + (1\hat{i} - 10\hat{j})(0.2) = 1.2\hat{i} + 0\hat{j} \text{ m/s}$. Position $y(0.2) = u_y t + \frac{1}{2} a_y t^2 = 2(0.2) - 5(0.2)^2 = 0.4 - 0.2 = 0.2 \text{ m}$.
• Motion for $t > 0.2$: Magnetic field $\vec{B} = 6\hat{j}$. Since gravity is also in $\hat{j}$, the vertical motion ($y$) is independent of the magnetic force ($\vec{v} \times \vec{B}$ has no $\hat{j}$ component). $y(t) = y(0.2) + v_y(0.2)(t-0.2) - 5(t-0.2)^2 = 0.2 - 5(t-0.2)^2$. At $t = 0.3$, $y = 0.2 - 5(0.1)^2 = 0.15 \text{ m} = 15 \text{ cm}$. (A) is correct. At $t = 0.4$, $y = 0.2 - 5(0.2)^2 = 0 \text{ cm}$. (B) is incorrect.
• Radius of Trajectory: The velocity $\vec{v}(0.2) = 1.2\hat{i}$ is perpendicular to $\vec{B} = 6\hat{j}$. $R = \frac{mv}{qB} = \frac{1.2}{1 \times 6} = 0.2 \text{ m} = 20 \text{ cm}$. (C) is correct.
Step 4: Final Answer:
The correct options are (A) and (C).
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