Two solenoidal coils, P and Q , are connected in different circuits. Coil Q is wound with twice the total number of turns as coil P , has twice the length, and half the cross-sectional area of coil P . If the current flowing through coil Q is twice that in coil P , the ratio of the magnetic energy stored in coil P to that in coil Q ( U_P : U_Q ) is

Step-by-step solution

The self-inductance of a solenoid is given by L = ₀ N^2 A l . Let the parameters for coil P be N_P , l_P , and A_P . For coil Q , we are given N_Q = 2N_P , l_Q = 2l_P , and A_Q = A_P 2 . The ratio of their self-inductances is: L_Q L_P = ( N_Q N_P )^2 ( A_Q A_P ) ( l_P l_Q ) L_Q L_P = (2)^2 ( 1 2 ) ( 1 2 ) = 4 1 4 = 1 Thus, L_P = L_Q . The magnetic energy stored in an inductor is U = 1 2 L i^2 . Since L is the same for both coils, U i^2 . Given that i_Q = 2i_P , the ratio of stored energy is: U_P U_Q = ( i_P i_Q )^2