Problem 1: The magnetic field at one place on the earth's surface is B = 55 μT in strength and tilted 60° down from horizontal. An N = 200-turn coil having a diameter of D = 4.0 cm and a resistance of R = 2.0 2 is connected to a C = 1.0 μF capacitor (see Fig. 1). The coil is held in a horizontal plane and the capacitor is discharged. Then the coil is quickly rotated 180° so that the side that had been facing up is now facing down. Afterward, what is the voltage across the capacitor? D 60° LA PI → a) When the coil is flipped, the magnetic flux through it is going to change, which will cause an induced current, which in turn will charge the capacitor. Start from computing the magnetic flux through the coil in its ini- tial state, assuming that the area vector is pointing down as shown in the scheme. After the coil is flipped upside down, the magnetic flux will change its sign to the opposite (think about flipping the area vector). What is the total change in magnetic flux, Am, in terms of N, B, D, and the relevant angle. FIG. 1: The scheme for Prob- lem 1 b) The induced current will bring the charge to the capacitor plates at the rate I = dq/dt. Using this and the formula for the induced emf, show that the charge accumulated at the capacitor plates immediately after the capacitor is flipped is related to the total change in magnetic flux as Δq = ΔΦm/R.
Problem 1: The magnetic field at one place on the earth's surface is B = 55 μT in strength and tilted 60° down from horizontal. An N = 200-turn coil having a diameter of D = 4.0 cm and a resistance of R = 2.0 2 is connected to a C = 1.0 μF capacitor (see Fig. 1). The coil is held in a horizontal plane and the capacitor is discharged. Then the coil is quickly rotated 180° so that the side that had been facing up is now facing down. Afterward, what is the voltage across the capacitor? D 60° LA PI → a) When the coil is flipped, the magnetic flux through it is going to change, which will cause an induced current, which in turn will charge the capacitor. Start from computing the magnetic flux through the coil in its ini- tial state, assuming that the area vector is pointing down as shown in the scheme. After the coil is flipped upside down, the magnetic flux will change its sign to the opposite (think about flipping the area vector). What is the total change in magnetic flux, Am, in terms of N, B, D, and the relevant angle. FIG. 1: The scheme for Prob- lem 1 b) The induced current will bring the charge to the capacitor plates at the rate I = dq/dt. Using this and the formula for the induced emf, show that the charge accumulated at the capacitor plates immediately after the capacitor is flipped is related to the total change in magnetic flux as Δq = ΔΦm/R.
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter22: Magnetic Forces And Magnetic Fields
Section: Chapter Questions
Problem 51P
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Hello, I am really need help with part A,Part B,Part C because I did the problem three times and I don't know why I get the wrong answer is there any way you can help me with this and can you label it as well. Thank you
Transcribed Image Text:Problem 1: The magnetic field at one place on the earth's surface is B =
55 μT in strength and tilted 60° down from horizontal. An N = 200-turn coil
having a diameter of D = 4.0 cm and a resistance of R = 2.0 2 is connected
to a C = 1.0 μF capacitor (see Fig. 1). The coil is held in a horizontal plane
and the capacitor is discharged. Then the coil is quickly rotated 180° so that
the side that had been facing up is now facing down. Afterward, what is the
voltage across the capacitor?
D
0
OPI
a) When the coil is flipped, the magnetic flux through it is going to
change, which will cause an induced current, which in turn will charge the
capacitor. Start from computing the magnetic flux through the coil in its ini-
tial state, assuming that the area vector is pointing down as shown in the scheme. After the coil is flipped
upside down, the magnetic flux will change its sign to the opposite (think about flipping the area vector).
What is the total change in magnetic flux, APm, in terms of N, B, D, and the relevant angle.
FIG. 1: The scheme for Prob-
lem 1
b) The induced current will bring the charge to the capacitor plates at the rate I = dq/dt. Using this and
the formula for the induced emf, show that the charge accumulated at the capacitor plates immediately
after the capacitor is flipped is related to the total change in magnetic flux as Δq = ΔΦm/R.
c) Use the capacitance C and the charge Aq to compute the voltage across the capacitor immediately
after it is flipped. (Answer: 12 V.)
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