Question

1. Starting with the expression for the magnetic field of a long solenoid with height h, radius R, and N turns, and ignoring end effects, derive the inductance L of such a solenoid using the relation Vz-LI. 2. Two coaxial loops of wire, of radii a, and a, are parallel to each other and separated by distance r> a, ,a2 along their common axis. The loops carry steady currents of I, and l2, Derive the mutual inductance M of the loops using either of the relations q, Mii, or ??-M12.

Answer 1

Self Inductance of a Solenoid The diagram shows an ideal solenoid with Area = A = pi r^2 radius r and the length of the solenoid is l > > r ie length is very high compared to the radius I is the current flowing through the solenoid. So the magnetic induction inside the solenoid is B = mu_0 (N/l)I - (1) Magentic flux through any cross section of solenoid is Phi_s = B. A = mu_0 NIA/l - (2) The total flux through the N turns of the solenoid is Phi = Phi_s times N = (mu_0 N^2 A/l)I - (3) In the question it is given that Phi = LI - (4) \r\n So we can write from equation (3) and equation(4) L = mu_0 N^2 A/l So L = mu_0 N^2 A/l is the self inductance of a solenoid. \r\n The magnetic flux produced by a variable current in the coil (primary) links with a closely situated coil (secondary), there is said to mutual induction If I_1 and I_2 are the current flowing through two coils then we can write Phi_1 = MI_2 & Phi_2 = MI_1 Now, the magnetic field due to current I_2 can be written as