Problem 3: An I = 10 A current is charging a D = 1.0 cm diameter parallel-plate capacitor. What is the magnetic field strength at 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 B. ds = Holthrough + oHo doe where B-ds is the line integral of magnetic field along a closed loop, Ithrough is the current passing through the loop, and Þe is the electric flux through the surface bounded by the loop. OO 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 value in each point of the loop, so f Bout ds = Bout f ds = 2лsBout ('out' 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 time. The rate at which the charge is supplied to the capacitor plates it I = dq/dt, so the charge at the plates as a function of time is q = f Idt = It (here we assume that the current is constant and q = 0 at t = 0). Use this to compute the electric flux De through a s = 2.0 mm-radius loop placed between the capacitor plates (as shown in Fig. 4) as a function of time. Bout FIG. 3: The scheme for Problem 3a changing electric field generates magnetic field in this region обо FIG. 4: The scheme for Problem 3b 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 Bin ds = Bin f ds = 27sBin. Note that the value of Bin at the distances 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 computed in the previous step. 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: Bin = 1.6 × 10-4 T.)

Problem 3: An I = 10 A current is charging a D = 1.0 cm diameter parallel-plate capacitor. What is the magnetic field strength at 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 B. ds = Holthrough + oHo doe where B-ds is the line integral of magnetic field along a closed loop, Ithrough is the current passing through the loop, and Þe is the electric flux through the surface bounded by the loop. OO 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 value in each point of the loop, so f Bout ds = Bout f ds = 2лsBout ('out' 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 time. The rate at which the charge is supplied to the capacitor plates it I = dq/dt, so the charge at the plates as a function of time is q = f Idt = It (here we assume that the current is constant and q = 0 at t = 0). Use this to compute the electric flux De through a s = 2.0 mm-radius loop placed between the capacitor plates (as shown in Fig. 4) as a function of time. Bout FIG. 3: The scheme for Problem 3a changing electric field generates magnetic field in this region обо FIG. 4: The scheme for Problem 3b 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 Bin ds = Bin f ds = 27sBin. Note that the value of Bin at the distances 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 computed in the previous step. 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: Bin = 1.6 × 10-4 T.)