Question

Exercise 30.4 Part A A solenoidal coil with 30 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 25.0 cm long and has a diameter of 2.00 cm. At a certain time, the current in the inner solenoid is 0.120 A and is increasing at a rate of 1500 A/S. For this time, calculate the average magnetic flux through each turn of the inner solenoid. Express your answer in webers. ADD A O O ? Submit Request Answer Part B For this time, calculate the mutual inductance of the two solenoids. Express your answer in henries. IVO AQ R O a ? M =
Please solve for Parts A,B and C

Answer 1

Ansuler "Given that Nr. 30 N -300 I = 25cm d=2cm 12 0.120 A = 1500 A/s. (a) me know that The magnetic field at the centery the solenold is B = Moni= Mon B, – umX107x300X0.120 0,25 ..By 1,81X1047 The flux through each turn of the Inner solenoid is PB = B; A= Barr 8B = 4 B, ad PB = xl.Bixio X 0,02 DB = 5.68 X16 8 wb I

b) we know that The magnetic flux through each turn of both solenoids is the same M= N₂ OBI – N₂ DB). (C) M- 30x 5.68x10? 01120 | M=1,42x105 H The emf induced in the outer solenoid due to changing current in the inner Solenoid Ez =- Mdi at C-142x10 5 x 1500 E = -0.0213 v