Question

A small, 1.90 mm-diameter circular loop with R= 1.40x 10-2 is at the center of a large 100 mm- diameter circular loop. Both loops lie in the same plane. The current in the outer.loop changes from +1.0 Ato -1.0 A in 8.00x10-2 s. Part A What is the induced current in the inner loop? Express your answer with the appropriate units. O': A OB? Value Units Submit Request Answer Provide Feedback

Answer 1

Solution The magnitude of the magnetic field at of the a large circular loop is given by the center where, i= r = current in the loop radius of the loop Therefore, the magnetic flux through a small circular loop is given by of - BA - (2) where, A= area of the small loop Now, since, A=TV12 where, r'-radius of the small loop Therefore, from equation (2), we have & = Briv2 Now, from equation (1), we get - (3) o = win82 => db = 7 M 72 ; - (4)

induced Now, the given by current in the immer loop is im =- & ad m = - 1 4 12 i – (5) Now, since, r= 100mm = 50mm = 50x103 m V = 1.90 mm = 0.95 mm = 0.95 x103 m R = 1.40x10² r, M= 4rx10 HIM si = – 1.0-(1.0) = -1.0-1.0 = - 2.0 A and ot = 8.00x1028

Therefore, we have from equation (5) as im = - 3.14 x 4x3.14x10 x 10.95x103) 2/-20 1 1.40x10? 2 150x103) (8.00x10² = m = 0.068106 A = 0.06 NA = lim = 0.06 WA Therefore, the induced current in the inner loop is lin = 0.06 M A