The Lagrangian of a particle of mass \( m \) and charge \( q \) moving in a uniform magnetic field of magnitude \( 2B \) that points in the \( z \)-direction, is given by: \[ L = \frac{m}{2} v^2 + qB(x v_y - y v_x) \] where \( v_x, v_y, v_z \) are the components of its velocity \( v \). If \( p_x, p_y, p_z \) denote the conjugate momenta in the \( x, y, z \)-directions and \( H \) is the Hamiltonian, which of the following option(s) is/are correct?
1. Conjugate momenta:
The conjugate momenta are given by: \[ p_x = \frac{\partial L}{\partial v_x} = m v_x + q B y \] \[ p_y = \frac{\partial L}{\partial v_y} = m v_y - q B x \] \[ p_z = \frac{\partial L}{\partial v_z} = m v_z \] 2. Hamiltonian:
The Hamiltonian \( H \) is the Legendre transform of the Lagrangian: \[ H = p_x v_x + p_y v_y + p_z v_z - L \] Substituting for \( v_x, v_y, v_z \) and simplifying the expression, we get: \[ H = \frac{1}{2m} \left[ (p_x + q B y)^2 + (p_y - q B x)^2 + p_z^2 \right] \] which matches option (D).
3. Equations of motion:
The equations of motion follow from the Hamiltonian dynamics. For \( \frac{d x}{d t} \), we have: \[ \frac{d x}{d t} = \frac{\partial H}{\partial p_x} = \frac{1}{m} (p_x - q B y) \] which matches option (A). Similarly, for \( \frac{d p_x}{d t} \), we get: \[ \frac{d p_x}{d t} = -\frac{\partial H}{\partial x} = \frac{q B}{m} (p_y - q B x) \] which matches option (B). For \( \frac{d p_y}{d t} \), we get: \[ \frac{d p_y}{d t} = -\frac{\partial H}{\partial y} = -\frac{q B}{m} (p_x + q B y) \] which matches option (C). Thus, the correct answers are (B), (C), and (D).
A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of \( 2 \times 10^5 \, \text{m/s} \). When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of electric field is \( x \times 10^4 \, \text{N/C} \). The value of \( x \) is \(\_\_\_\_\_\). (Take the mass of the proton as \( 1.6 \times 10^{-27} \, \text{kg} \)).
Due to presence of an em-wave whose electric component is given by \( E = 100 \sin(\omega t - kx) \, NC^{-1} \), a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as:
In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
The figure shows an opamp circuit with a 5.1 V Zener diode in the feedback loop. The opamp runs from \( \pm 15 \, {V} \) supplies. If a \( +1 \, {V} \) signal is applied at the input, the output voltage (rounded off to one decimal place) is:
A wheel of mass \( 4M \) and radius \( R \) is made of a thin uniform distribution of mass \( 3M \) at the rim and a point mass \( M \) at the center. The spokes of the wheel are massless. The center of mass of the wheel is connected to a horizontal massless rod of length \( 2R \), with one end fixed at \( O \), as shown in the figure. The wheel rolls without slipping on horizontal ground with angular speed \( \Omega \). If \( \vec{L} \) is the total angular momentum of the wheel about \( O \), then the magnitude \( \left| \frac{d\vec{L}}{dt} \right| = N(MR^2 \Omega^2) \). The value of \( N \) (in integer) is: