Q20: A square loop of copper wire is pulled through a... - Electromagnetic Induction and Alternating Currents - Physics CUET UG 02-07-2023 Shift 3 Paper | Agrigyan

Analysis of the four positions of the loop:

• Position A (Loop entirely outside the magnetic field): Magnetic flux $\Phi = 0$, so induced emf $\mathcal{E} = 0$, current $I = 0$, and magnetic force $F_{\text{mag}} = 0$. Thus, pulling force $F_A = 0$.
• Position B (Loop entering the field): Magnetic flux is increasing at rate $\frac{d\Phi}{dt} = B l v$. Induced current $I = \frac{B l v}{R}$ experiences a magnetic retarding force $F_{\text{mag}} = I l B = \frac{B^2 l^2 v}{R}$. To maintain constant speed, $F_B = \frac{B^2 l^2 v}{R}$.
• Position C (Loop entirely inside the uniform field): Magnetic flux is maximum and constant ($\frac{d\Phi}{dt} = 0$). No induced emf or current exists, so $F_{\text{mag}} = 0$. Thus, pulling force $F_C = 0$.
• Position D (Loop exiting the field): Magnetic flux is decreasing at rate $\frac{d\Phi}{dt} = -B l v$. Induced current produces an opposing retarding force $F_{\text{mag}} = \frac{B^2 l^2 v}{R}$. To maintain constant speed, $F_D = \frac{B^2 l^2 v}{R}$.
  
  Comparing magnitudes:
  $$F_B = F_D = \frac{B^2 l^2 v}{R} \quad \text{and} \quad F_A = F_C = 0$$