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24. (a) We assume the flux is entirely due to the field generated by the long straight wire (which is given by Eq. 29-17). We integrate according to Eq. 30-1, not worrying about the possibility of an overall minus sign since we are asked to find the absolute value of the flux. µ 0ia § r + b2 · § µ 0i · a dr = ( ) ln ¨ . b ¸ r − b / 2 ¨ 2πr ¸ 2π © ¹ ©r−2¹ | Φ B |= ³ r +b / 2 When r = 1.5b , we have | ΦB | = (4π × 10 −7 T ⋅ m A)(4.7A)(0.022m) ln(2.0) = 1.4 ×10−8 Wb. 2π (b) Implementing Faraday’s law involves taking a derivative of the flux in part (a), and recognizing that drdt = v . The magnitude of the induced emf divided by the loop resistance then gives the induced current: iloop = ε R =− µ 0ia d § r + b2 · µ 0iabv = ln ¨ b ¸ 2πR dt © r − 2 ¹ 2πR[r 2 − (b / 2) 2 ] (4π ×10 −7 T ⋅ m A)(4.7A)(0.022m)(0.0080m)(3.2 ×10−3 m/s) = 2π (4.0 ×10−4 Ω)[2(0.0080m) 2 ] = 1.0 ×10−5 A.