Problem 3: An I = 10 A current is charging a D = 1.0 cm diameter parallel-plate capacitor. What is the magnetic field strength al a point s = 2.0 mm radially from the center of the wire leading to the capacitor? What is the magnetic field strength at a point s = 2.0 mm radially from the center of the capacitor? a) Ampere-Maxwell law reads f 8 - d5 = Palthough + Calo of. , where f B-ds is the line integral of magnetic field along a closed loop. through is the current passing through the loop, and $, is the electric flux through the surface bounded by the loop. In Fig 3, we place a loop of radius s = 2.0 mm around the current leading to the capacitor. The magnetic field created by the current is circumferential, and due to the symmetry of the problem it has the same FIG. 3: The scheme for Problem 3a value in each point of the loop, so f Bou-d5 = Be f ds = 2xsham Coul" is for outside the capacitor). What is the current passing through the loop? The electric flux through this loop is equal to zero - can you explain why? Use this to compute the magnetic field strength at 2.0 mm from the center of the wire. b) Inside the capacitor, the electric field is changing with lime. The changing cientin field rate at which the charge is supplied to the capacitor plates it I = dy/di, generates mogmelin field be this region so the charge at the plates as a function of time is q = / Idi = It (here we assume that the current is constant and q = 0 alf = 0). Use this to compute the electric flux +: through a s = 2.0 mm-radius loop placed between the capacitor plates (as shown in Fig. 4) as a function of time. Bim FIG. 4: The scheme for Problem ib c) The scheme in Fig. 4 is meant to emphasize that the changing electric field between the plates of the capacitor creates circumferential magnetic field similar to the one of a current. Placing a symmetric loop of radius s = 2.0 mm between the plates, we get for the line integral of the magnetic field inside the capacitor f Bin - ds = Bif ds = 2xsBy. Note that the value of Be, at the distance s from the symmetry axis is not the same as Bout, but it is the same at each point along the loop, and that is sufficient to simplify the line integral. In Fig. 4 there is no physical current through the loop, but there is a changing electric flux, which you compuled in the previous slep. Use Ampere-Maxwell law to compute the magnetic field strength at a point s = 2.0 mm radially from the center of the capacitor? (Answer: Be, = 1.6 x 10 'T.)