30.4fill in blanks Ed-(BA cos 0) =dtdI= -(-²) (HONA) Iкноп dtε = -(²-) (μ₁NA) DIdtddt2Note that the area A is interpreted as the area A = ₂² of the solenoid, where the field is strong. Substituting,we haveII(MonIA cos 0°)HONIA= −0.500(4 × 10¯7 T · m/A) (1= (-955x 10-4 V.V. Cx 10−4 T · m²/s)( 1. N. 5) (1¹.C.m. T• mBsolenoidturns/m) (ISTm)² (2712 ==An aluminum ring of radius ₁ 5.00 cm and a resistance of 2.50 × 10-4 is placed around one end of a longair-core solenoid with 955 turns per meter and radius 2 3.00 cm as shown in the figure below. Assume theaxial component of the field produced by the solenoid is one-half as strong over the area of the end of thesolenoid as at the center of the solenoid. Also assume the solenoid produces negligible field outside its cross-sectional area. The current in the solenoid is increasing at a rate of 270 A/s.↑71(a) What is the induced current in the ring?(b) At the center of the ring, what is the magnitude of the magnetic field produced by the induced currentin the ring?(c) At the center of the ring, what is the direction of the magnetic field produced by the induced currentin the ring?

Answered: Image /qna-images/answer/43560299-9dc2-430e-a91c-77ff9ccc9326.jpg

ε = - (BA cos 0) dt ε = -- () (MONA) 71 dI -(-) (MORA) dt = Note that the area A is interpreted as the area A we have MONIA dI dt = -0.500(4 x 10-7 T. m/A) ( 955 -(- Cos 0°) x 10-4 V. Bsolenoid 127 2 of the solenoid, where the field is strong. Substituting, x 10-4 T + m²/s) 1.N-S-¹V-C) c.m.T) ( ₂ turns/m) ( T m)² (270 A/

ε = - (BA cos 0) dt ε = -- () (MONA) 71 dI -(-) (MORA) dt = Note that the area A is interpreted as the area A we have MONIA dI dt = -0.500(4 x 10-7 T. m/A) ( 955 -(- Cos 0°) x 10-4 V. Bsolenoid 127 2 of the solenoid, where the field is strong. Substituting, x 10-4 T + m²/s) 1.N-S-¹V-C) c.m.T) ( ₂ turns/m) ( T m)² (270 A/