Example 31.8 Electric Field Induced by a Changing Magnetic Field in a Solenoid P

Question

Example 31.8 Electric Field Induced by a Changing Magnetic Field in a Solenoid 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 1-Imax cos ot, where Imax is the maximum current and is the angular frequency of the AC source (Fig. 31.20) Path of Integration Determine the magnitude of the induced electric field outside the solenoid, a distance r > R from its long central axis B What is the magnitude of the induced electric field inside the solenoid, a distance r from its axis? max cos ot Figure 31.20 A long solenoid carrying a time-varying current given by 1 = Imax cos ot. An electric field is induced both 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 magniude of E is constant on this path and that E is tangent to it. The magnetic flux through the area enclosed by the path is Solution Equation 31.9 gives the following dt dt The magnetic field inside a long solenoid is given by Equation 30.17, BonI. When we substitute the expresssion IImax cos t into this equation for B and then substitute the result into Equation (1), we find the following (use mu-0 for po, 1 for Imax, omega for , and n, r, R, and t as necessary) dt (2) E (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

Explanation / Answer

The integral of the electric field around a closed loop = the rate of change of magnetic flux thru the loop.

For constant area the rate of change of flux is given by
d(BA)/dt = A*dB/dt

Now, Imagine a hypothetical circle with a radius equal to the value you want to know E at, i.e 1.67 cm

Therefore we can write -
Integral(E.dl) = -AdB/dt

P.S - minus sign just shows the direction of E (clockwise or counterclockwise).

Integral(E dot dl) = E(2)(pi)(r)
Also, A = (pi)(r^2)

Then

E = (1/2)(r)dB/dt
E = (1/2)(R)* d/dt(0.04t^2 + 2.95)
E = (1/2)(.0167)[(.08)t]

Substituing, t = 7.5 sec;
E = (1/2)(.0167)(.08)(7.5)
E = .00501 N/C

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