Problem#1
The magnetic field within a long, straight solenoid with a circular cross section and radius R is increasing at a rate of dB/dt. (a) What is the rate of change of flux through a circle with radius r1 inside the solenoid, normal to the axis of the solenoid, and with center on the solenoid axis? (b) Find the magnitude of the induced electric field inside the solenoid, at a distance r1 from its axis. Show the direction of this field in a diagram. (c) What is the magnitude of the induced electric field outside the solenoid, at a distance from r2 the axis? (d) Graph the magnitude of the induced electric field as a function of the distance from r the axis from r = 0 to r = 2R. (e) What is the magnitude of the induced emf in a circular turn of radius R/2 that has its center on the solenoid axis? (f) What is the magnitude of the induced emf if the radius in part (e) is R? (g) What is the induced emf if the radius in part (e) is 2R?Answer:
(a) the rate of change of flux through a circle with radius r1 inside the solenoid, normal to the axis of the solenoid, and with center on the solenoid axis is given by
dΦB/dt = d(B.A)/dt = πr12 (dB/dt)
(b) the magnitude of the induced electric field inside the solenoid, at a distance r1 from its axis is given by
E = (1/2πr1)(dΦB/dt) = (1/2πr1)(πr12)(dB/dt)
E = (dB/dt) r1/2
The direction of E is shown in Figure 1a.
(c) the magnitude of the induced electric field outside the solenoid, at a distance from r2 the axis is (All the flux is within r < R),
E = (1/2πr2)(dΦB/dt) = (πR2/2πr2)(dB/dt)
E = (dB/dt) R2/2r2
(d) Graph the magnitude of the induced electric field as a function of the distance from r the axis from r = 0 to r = 2R in Figure 1b.
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| Fig.1 |
(e) the magnitude of the induced emf in a circular turn of radius R/2 that has its center on the solenoid axis is
dΦB/dt = π(R/2)2 (dB/dt) = (dB/dt)πR2/4
(f) the magnitude of the induced emf in a circular turn of radius R that has its center on the solenoid axis is
dΦB/dt = π(R)2 (dB/dt) = (dB/dt)πR2
(g) the magnitude of the induced emf in a circular turn of radius 2R that has its center on the solenoid axis is
dΦB/dt = (dB/dt)πR2
The emf is independent of the distance from the center of the cylinder at all points outside it. Even though the magnetic field is zero for r > R, the induced electric field is nonzero outside the solenoid and a nonzero emf is induced in a circular turn that has r > R.
Problem#2
A long, thin solenoid has 900 turns per meter and radius 2.50 cm. The current in the solenoid is increasing at a uniform rate of 60.0 A/s. What is the magnitude of the induced electric field at a point near the center of the solenoid and (a) 0.500 cm from the axis of the solenoid; (b) 1.00 cm from the axis of the solenoid?
Answer:
The end view of the solenoid is sketched in Figure 2. Let R be the radius of the solenoid.
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| Fig.2 |
∮ E.dL = E(2πr) and
|-dφB/dt| = πr2 |dB/dt|
Then,
E(2πr) = πr2 |dB/dt|
E = ½ r|dB/dt|
Solenoid magnetic field is
B = µ0nI, so dB/dt = µ0n(dI/dt)
Thus, E = ½ r|dB/dt| = ½rµ0n(dI/dt)
E = ½ x 0.0500 m x 4π x 10-7 T.m/A x 900/m x 60.0 A/s = 1.70 x 10-4 V/m
(b) 0.0100 cm r = is still inside the solenoid so the expression in part (a) applies.
E = ½ x 0.0100 m x 4π x 10-7 T.m/A x 900/m x 60.0 A/s = 3.39 x 10-4 V/m
Problem#3
A long, thin solenoid has 400 turns per meter and radius 1.10 cm. The current in the solenoid is increasing at a uniform rate di/dt. The induced electric field at a point near the center of the solenoid and 3.50 cm from its axis is 8.00 x 10-6 V/s. Calculate di/dt.
Answer:
induction emf is given by
ε = |-dφB/dt| = |-d(B.A)/dt|
with Solenoid magnetic field is
B = µ0nI, so dB/dt = µ0n(dI/dt)
Then
ε = |-dφB/dt| = µ0nA(dI/dt)
because
ε = E.(2πr)
µ0nA(dI/dt) = E.(2πr)
dI/dt = 2πrE/µ0nA
= 2π x 0.0350 m x (8.00 x 10-6 V/s)/(4π x 10-7 T.m/A x 400 x π x (0.0110 m)2)
dI/dt = 9.21 A/s
Outside the solenoid the induced electric field decreases with increasing distance from the axis of the solenoid.
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