Problem#1
A single loop of wire with an area of 0.0900 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.190 T/s. (a) What emf is induced in this loop? (b) If the loop has a resistance of find the current induced in the loop.Answer:
(a) An EMF induced in the loop which causes a current to flow in it. As
ε = dφB/dt
with φB = NBA cos θ, θ = 00 and N = 1.
A is constant and B is changing so,
ε = A (dB/dt) = (0.0900 m2)(0.190 T/s) = 0.0171 V
(b) the current induced in the loop is
I = ε/R = 0.0171 V/0.600 Ω = 0.0285 A
Problem#2
In a physics laboratory experiment, a coil with 200 turns enclosing an area of 12 cm2 is rotated in 0.040 s from a position where its plane is perpendicular to the earth’s magnetic field to a position where its plane is parallel to the field. The earth’s magnetic field at the lab location is 6.0 x 10–5 T. (a) What is the total magnetic flux through the coil before it is rotated? After it is rotated? (b) What is the average emf induced in the coil?
Answer:
(a) An EMF induced in the loop which causes a current to flow in it. As
ε = dφB/dt
with φB = BA cos θ, θi = 00, θf = 900 and N = 200.
When,
θi = 00, φB,i = 6.0 x 10–5 T x 12 x 10–4 m2 cos 00 = 7.20 x 10–8 Wb.
θf = 900, φB,f = 6.0 x 10–5 T x 12 x 10–4 m2 cos 900 = 0 Wb.
(b) the average emf induced in the coil is
ε = –NΔφB/Δt = (φB,f – φB,i)/Δt
ε = –200(0 – 7.20 x 10–8 Wb)/(0.040 s) = +0.36 mV
Problem#3
One practical way to measure magnetic field strength uses a small, closely wound coil called a search coil. The coil is initially held with its plane perpendicular to a magnetic field. The coil is then either quickly rotated a quarter-turn about a diameter or quickly pulled out of the field. (a) Derive the equation relating the total charge Q that flows through a search coil to the magnetic-field magnitude B. The search coil has N turns, each with area A, and the flux through the coil is decreased from its initial maximum value to zero in a time Δt. The resistance of the coil is R, and the total charge is Q = IΔt where I is the average current induced by the change in flux. (b) In a credit card reader, the magnetic strip on the back of a credit card is rapidly “swiped” past a coil within the reader. Explain, using the same ideas that underlie the operation of a search coil, how the reader can decode the information stored in the pattern of magnetization on the strip. (c) Is it necessary that the credit card be “swiped” through the reader at exactly the right speed? Why or why not?
Answer:
(a) the magnitude of the emf induced in the coil is
|εav| = N|ΔΦB/Δt| = N[ΦBf – ΦBi]/Δt
With, ΦB = BA cos θ
Initially (θ = 0), ΦBi = BA cos θ = BA
The final flux is zero. Because θ = 900, ΦBf = BA cos 900 = 0
So,
εav = NBA/Δt
the average induced current is
I = |εav|/R = NBA/RΔt
The total charge that flows through the coil is
Q = IΔt
Q = NBA/R
(b) the magnetic stripe consist of a pattern of magnetic fields. The pattern of changes that flows in the reader coil.
(c) according to the result in part (a) the change that flows depends only on the change in the magnetic flux and it does not depend on the rate at which this flux changes.
Problem#4
A closely wound search coil (see Problem#3) has an area of 3.20 cm2, 120 turns, and a resistance of 60.0 Ω. It is 45 Ω connected to a charge-measuring instrument whose resistance is When the coil is rotated quickly from a position parallel to a uniform magnetic field to a position perpendicular to the field, the instrument indicates a charge of 3.56 x 10-5 C. What is the magnitude of the field?
Answer:
Apply the result derived in problem#3:
Q = NBA/R
In present exercise the flux changes from its maximum value of ΦB = BA to zero, so this equation applies.
R is the total resistance so here
R = 60.0 Ω + 45.0 Ω = 105.0 Ω
So that
B = QR/NA
B = (3.56 x 10-5 C)(105.0 Ω)/(120 x 3.20 x 10-4 m2)
B = 0.0973 T
Post a Comment for "Faraday’s Law Problems and Solutions"