Problem A long solenoid of radius R has n turns of wire per unit length and carries a time-varying current that varies sinusoidally as I = Imax cos wt, where Imax is the maximum current and w is the angular frequency of the AC source (Fig. 31.20). A Determine the magnitude of the induced electric field outside the solenoid, a distance r> R from its long central axis. Bwhat is the magnitude of the induced electric field inside the solenoid, a distance r from its axis? Path of Integration max cos cot Figure 31.20 A long solenoid carry time-varying current given by Ir cos wt. An electric field is induced b inside and outside the solenoid. Strategy First consider an external point and take the path for our line integral to be a circle of radius r centered on the solenoid as illustrated in Figure 31.20. By symmetry, we see that the magnitude of E is constant on this path and that E is tangent to it. The magnetic flux through the area enclosed by the path ia = BRR2, Solution A Equation 31.9 gives the following. d E-ds--- (B=R²)=-RB fEd (1) E.. dt E ds = E(2x)=-R dt dt The magnetic field inside a long solenoid is given by Equation 30.17, B = Hon. When we substitute the expresssion IImax cos wt into this equation for B and then substitute the result into Equation (1), we find the following (Use the following as necessary: Ho. Imax, w, n, r, R, and t.) d E (2xr) --Ron Imax) cos wt dt R2 2r (2) E-ma Ron Imaxwsin wt -sin(cot) (for r > R) Hence, the electric field varies sinusoidally with time, and its amplitude falls off as 1/r outside the solenoid. According to the Ampère-Maxwell law, the changing electric field creates an additional contribution to the magnetic field. At high frequencies, an altogether new phenomenon can occur. The electric and magnetic fields, each supporting the other, can constitute an electromagnetic wave radiated by the solenoid. For an interior point (r<R), the flux passing through the area bounded by a path of integration is given by Br2, Using the same procedure as in part (A), we find the following (Use the following as necessary: Jo. Imax w, n, r, and t.) E (2r) B dt r²on Imax sin t O Type here to search 4 C Home | Chegg.com x G If the resistance in the cox N230 HW Sb Faraday's X C1)A Flat Loop Of Wire Cx Cwww.webassign.net/web/Student/Assignment-Responses/submit/dep-18089352 J dt E ds E(2r)=-1 dt dr (1) The magnetic field inside a long solenoid is given by Equation 30.17, B = Hon. When we substitute the expresssion I Imax cos out into this equation for B and then substitute the result into Equation (1), we find the following (Use the following as necessary: Ho. Imax, w, n, r, R, and t.) d Ronmaxcost-pon Imax sin f E(2) (2) E-M 2r -sin (cor) (for r>R) Hence, the electric field varies sinusoidally with time, and its amplitude falls off as 1/r outside the solenoid. According to the Ampère-Maxwell law, the changing electric field creates an additional contribution to the magnetic field. At high frequencies, an altogether new phenomenon can occur. The electric and magnetic fields, each supporting the other, can constitute an electromagnetic wave radiated by the solenoid. For an interior point (< R), the flux passing through the area bounded by a path of integration is given by Ber. Using the same procedure as in part (A), we find the following (Use the following as necessary: Ho. Imax w, n, r, and t.) E (2) B pon/max sin f dt (3) E- max 2 -rsin(cot) (for r<R) This expression shows that the amplitude of the electric field induced inside the solenoid by the changing magnetic field varies linearly with r and varies sinusoidally with time. Exercise 31.8 Hints: Getting Started | I'm Stuck A magnetic field directed along the x-axis changes with time according to B- (0.052 3.95) T, where t is in seconds. The field is confined to a circular beam of radius 3.85 cm. What is the magnitude of the electric field at a point 1.99 cm measured perpendicular from the x-axis when t5.50 s? E= N/m -4 points PSE6 31 AF 10