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GATE 2025 PH – Question 51

Classical Mechanics · Hamiltonian and Hamilton's equations of motion · 2 marks · Multiple select

A particle of mass $m$ and charge $q$ in uniform magnetic field $2B\hat z$ has $L=\frac m2v^2+qB(xv_y-yv_x)$. With conjugate momenta $p_x,p_y,p_z$ and Hamiltonian $H$, which option(s) is/are correct?

  1. $\dot x=(p_x-qBy)/m$
  2. $\dot p_x=qB(p_y-qBx)/m$
  3. $\dot p_y=-qB(p_x+qBy)/m$
  4. $H=[(p_x+qBy)^2+(p_y-qBx)^2+p_z^2]/(2m)$

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Show answer and explanation

Correct answer: (B) $\dot p_x=qB(p_y-qBx)/m$; (C) $\dot p_y=-qB(p_x+qBy)/m$; (D) $H=[(p_x+qBy)^2+(p_y-qBx)^2+p_z^2]/(2m)$

Explanation

**Momenta.** From $L=\tfrac m2v^2+qB(xv_y-yv_x)$:
$$p_x=mv_x-qBy,\qquad p_y=mv_y+qBx,\qquad p_z=mv_z .$$

**Velocities in terms of momenta:** $v_x=\dfrac{p_x+qBy}{m}$, $v_y=\dfrac{p_y-qBx}{m}$.
So option A ($\dot x=(p_x-qBy)/m$) has the wrong sign. ✗

**Hamiltonian** $H=\mathbf p\cdot\mathbf v-L=\tfrac m2v^2$ (the magnetic term does no work):
$$H=\frac{(p_x+qBy)^2+(p_y-qBx)^2+p_z^2}{2m}. ✓\ (\text{D})$$

**Hamilton's equations:**
$$\dot p_x=-\frac{\partial H}{\partial x}=-\frac{(p_y-qBx)(-qB)}{m}=\frac{qB(p_y-qBx)}{m}. ✓\ (\text{B})$$
$$\dot p_y=-\frac{\partial H}{\partial y}=-\frac{qB(p_x+qBy)}{m}. ✓\ (\text{C})$$

Answer **B, C and D**.