A coil is placed in a time varying magnetic field. The power dissipated due to current induced in the coil is P₁ . If the number of turns is doubled and radius of the wire is halved, the power dissipated is P₂ . Then P₁: P₂ is

Step-by-step solution

According to the question, radius of wire become r 2 , so its length will be 4 l and its resistance will become 16 R . The number of turns is doubled in a coil, so its radius should be doubled to accomodate the length of wire. The area of coil will become 4 times. Now, current induced in the coil is P , P₁= V₁^2 R From Eqs. (i) and (ii), we get gathered P₂= (8 V₁ )^2 16 R P₂= 64 V₁^2 16 R P₂= 4 V₁^2 R gathered Then the ratio, P₁: P₂= V₁^2 R : 4 V₁^2 R or P₁: P₂=1: 4