At \( t = 10s \), the complete loop is already inside the magnetic field. Since the loop is entirely within the magnetic field, there is no change in the flux as the loop moves through the field. The rate of change of magnetic flux \( \frac{d\Phi}{dt} \) is zero because the area of the loop inside the magnetic field remains constant.
Thus, the induced emf is:
\( e = \frac{d\Phi}{dt} = 0. \)
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:
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 \( 1.6 \times 10^{-27} \, \text{kg} \)).
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?
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:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: