Question

3 The magnetic field inside a toroidal solenoid is not uniform as for the long, straight solenoid. Over the cross-sectional area of the toroid the magnetic field is stronger near the inner HoNI radius of the torus and somewhat weaker near the outer radius according to B(r) 2Tr So we cannot technically use the simple expression P2 = BAfor the flux through one turn of wire. Nevertheless, the textbook uses an approximate constant value for an equivalent uniform magnetic field in a toroidal solenoid. The authors substitute the mean/average Iadius imesn of the toroid for the variable r. If the toroid has an inner radius of a and an b)/2 outer radius of b, then Tmean = (a a. Suppose a toroidal solenoid has a cross-sectional area of 6.25 cm2 and 500 turns of wire. If its self inductance is 0.7S1 mH, calculate its approximate mean radius b. If the current decreases from 5.84 A to 2.00 A in 3.00 ms, what is the induced voltage in this solenoid (to three significat figures)?

Answer 1

Criven (a) Az 6.25cm N = 500 turns. L20.781 mH = LEN(): N() R- MANIA NMA 2 TL Rz 500x 500 x 4xx107x 6.25 x 104 2AX 0 781810-3 Ra 0.4m b) 4Iz 5.84 A-200 A Z 3.84A in st z 3 ms. & dzbi at 3 E= -26 0) e dBA) 7:-d &z - L d I dt z-ZA -LAI E - 0.781 110 x 3.84 3 3x103 E e IV