Question

A solenoidal coil with 30 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 23.0 cm long and has a Part A diameter of 2.10 cm. At a certain time, the current in the inner solenoid is 0.120 A and is increasing at a rate of 1800 A/s. For this time, calculate the average magnetic flux through each turn of the inner solenoid. Express your answer in webers. Vη ΑΣφ' Фв - Wb Request Answer Submit Part B For this time, calculate the mutual inductance of the two solenoids. Express your answer in henries. αν ΑΣφ Н Submit Request Answer Part C For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid. Express your answer in volts. V ΑΣφ

Answer 1

Solution arven values Solenoid Cail torn N= oturns Another coil turn Nga 350 torn. Inner Solenoid long 2 23.0cm diameter 2 2.10cm radiusz 1.05cm 20.0105m I = 0.120A Increasing rate dI a 1800 Als dot Average magnetic flux magnetic field at the center of the solenoide Biz Moni - Mo NI I BI 4 ITX107 x 30 x 0.120 0.23 m (B12 1.968105 T magnetic flux Ø2 BA 02 B, 1T72 → 1-96x 165x314x(0.01052 1882 1.78X10 9 Wb to mutual Inductance of the two solenoide b) M= N2 dB Un OBI - Nga Bi > mutual Inductonce both etonce both sol solenoide sane

M2 N20 B 2 350 x1-784109 I Given 0120 N2 2 350 form M2.5.19 X 106 4B2 1-788159 wb TMz 5. 19x106H Calculate emf- Calculate solenoid due to changing The Induced Emf in the outer current in the innes solenoid. & unde dt Given diz a 1800 A/S M2 s. 19x1064 E - s19x106 & 1800 Ez - 0.60 93 v