Question:
Consider a particle of mass \(m = 9.0 \times 10^{-5}\) kg and charge \(q = 3.0 \times 10^{-4}\) C in a uniform electromagnetic field \(\vec{E} = 2\,\hat{x}\) V.m\(^{-1}\), \(\vec{B} = 3\,\hat{z}\) V.m\(^{-2}\).s. The particle is released from the coordinates \((0, 5\text{ m}, 0)\) at time \(t = 0\). Starting from initial speed zero, it comes back to the \(y\)-axis for the first time at time \(t\). The value of \(t\) in seconds (rounded off to two decimal places) is ______
Consider a particle of mass \(m = 9.0 \times 10^{-5}\) kg and charge \(q = 3.0 \times 10^{-4}\) C in a uniform electromagnetic field \(\vec{E} = 2\,\hat{x}\) V.m\(^{-1}\), \(\vec{B} = 3\,\hat{z}\) V.m\(^{-2}\).s. The particle is released from the coordinates \((0, 5\text{ m}, 0)\) at time \(t = 0\). Starting from initial speed zero, it comes back to the \(y\)-axis for the first time at time \(t\). The value of \(t\) in seconds (rounded off to two decimal places) is ______
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Hint:
A particle starting from rest in crossed \(\vec{E}\) and \(\vec{B}\) traces a cycloid: it returns to its starting \(x\)-line once every full cyclotron period \(2\pi/\omega\) with \(\omega = qB/m\). Solve the coupled \(v_x, v_y\) equations, or just find \(\omega\) and use this period directly.
A particle starting from rest in crossed \(\vec{E}\) and \(\vec{B}\) traces a cycloid: it returns to its starting \(x\)-line once every full cyclotron period \(2\pi/\omega\) with \(\omega = qB/m\). Solve the coupled \(v_x, v_y\) equations, or just find \(\omega\) and use this period directly.
Updated On: Jul 28, 2026
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Correct Answer: 0.63
Solution and Explanation
Step 1: Understand the setup.
The particle starts from rest at \((0, 5\text{ m}, 0)\), in a constant electric field \(\vec{E} = E\hat{x}\), \(E = 2\) V/m, and a constant magnetic field \(\vec{B} = B\hat{z}\), \(B = 3\) T (the unit V.m\(^{-2}\).s is just tesla written in base SI units). Crossed \(\vec{E}\) and \(\vec{B}\) fields, with the particle starting at rest, always produce a cycloid: the particle keeps coming back to its starting \(x\)-position (here, the \(y\)-axis) once every full cyclotron period.
Step 2: Write the equations of motion.
The Lorentz force is \(m\dot{\vec{v}} = q(\vec{E} + \vec{v}\times\vec{B})\). With \(\vec{B}=B\hat{z}\), \(\vec{v}\times\vec{B} = (v_yB, -v_xB, 0)\), so
\[ \dot{v}_x = \frac{qE}{m} + \omega v_y, \qquad \dot{v}_y = -\omega v_x, \qquad \omega = \frac{qB}{m} \]
Step 3: Solve the coupled equations.
Define \(u = v_y + E/B\) (using \(qE/m = \omega E/B\)); then \(\dot v_x = \omega u\) and \(\dot u = -\omega v_x\), a pure rotation in the \((v_x,u)\) plane. With \(v_x(0)=0\) and \(u(0) = E/B\) (since \(v_y(0)=0\)), the solution is
\[ v_x(t) = \frac{E}{B}\sin(\omega t), \qquad v_y(t) = \frac{E}{B}\big(\cos(\omega t) - 1\big) \]
Step 4: Integrate to get \(x(t)\).
\[ x(t) = \int_0^t v_x\,dt' = \frac{E}{B\omega}\big(1-\cos(\omega t)\big) \]
This is the coordinate away from the starting line \(x=0\) (the \(y\)-axis). The particle returns to the \(y\)-axis whenever \(x(t)=0\) again, that is \(\cos(\omega t) = 1\), which first happens after \(t=0\) at \(\omega t = 2\pi\).
Step 5: Compute \(\omega\) and \(t\).
\[ \omega = \frac{qB}{m} = \frac{(3.0\times10^{-4})(3)}{9.0\times10^{-5}} = 10 \text{ rad/s} \]
\[ t = \frac{2\pi}{\omega} = \frac{2\pi}{10} = 0.6283 \text{ s} \]
Final Answer:
Rounded to two decimal places, the particle first returns to the \(y\)-axis at \[ \boxed{t = 0.63 \text{ s}} \]
The particle starts from rest at \((0, 5\text{ m}, 0)\), in a constant electric field \(\vec{E} = E\hat{x}\), \(E = 2\) V/m, and a constant magnetic field \(\vec{B} = B\hat{z}\), \(B = 3\) T (the unit V.m\(^{-2}\).s is just tesla written in base SI units). Crossed \(\vec{E}\) and \(\vec{B}\) fields, with the particle starting at rest, always produce a cycloid: the particle keeps coming back to its starting \(x\)-position (here, the \(y\)-axis) once every full cyclotron period.
Step 2: Write the equations of motion.
The Lorentz force is \(m\dot{\vec{v}} = q(\vec{E} + \vec{v}\times\vec{B})\). With \(\vec{B}=B\hat{z}\), \(\vec{v}\times\vec{B} = (v_yB, -v_xB, 0)\), so
\[ \dot{v}_x = \frac{qE}{m} + \omega v_y, \qquad \dot{v}_y = -\omega v_x, \qquad \omega = \frac{qB}{m} \]
Step 3: Solve the coupled equations.
Define \(u = v_y + E/B\) (using \(qE/m = \omega E/B\)); then \(\dot v_x = \omega u\) and \(\dot u = -\omega v_x\), a pure rotation in the \((v_x,u)\) plane. With \(v_x(0)=0\) and \(u(0) = E/B\) (since \(v_y(0)=0\)), the solution is
\[ v_x(t) = \frac{E}{B}\sin(\omega t), \qquad v_y(t) = \frac{E}{B}\big(\cos(\omega t) - 1\big) \]
Step 4: Integrate to get \(x(t)\).
\[ x(t) = \int_0^t v_x\,dt' = \frac{E}{B\omega}\big(1-\cos(\omega t)\big) \]
This is the coordinate away from the starting line \(x=0\) (the \(y\)-axis). The particle returns to the \(y\)-axis whenever \(x(t)=0\) again, that is \(\cos(\omega t) = 1\), which first happens after \(t=0\) at \(\omega t = 2\pi\).
Step 5: Compute \(\omega\) and \(t\).
\[ \omega = \frac{qB}{m} = \frac{(3.0\times10^{-4})(3)}{9.0\times10^{-5}} = 10 \text{ rad/s} \]
\[ t = \frac{2\pi}{\omega} = \frac{2\pi}{10} = 0.6283 \text{ s} \]
Final Answer:
Rounded to two decimal places, the particle first returns to the \(y\)-axis at \[ \boxed{t = 0.63 \text{ s}} \]
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