Tuesday, July 8th - Notes Prepared by Melanie Smith
Quantitative Problem #2 from Yesterday’s Discussion, July 7th
Clarence borrowed from Essentials of College Physics, p 542 #14)
A circular coil enclosing an area of 100 sq. cm is made of 200 turns of copper wire. The
wire making up the coil has resistance of 5.0 ohms, and the ends of the wire are connected to
form a closed circuit. Initially, a 1.1-T uniform magnetic field points perpendicularly upward
through the plane of the coil. The direction of the field then reverses so that the final magnetic
field has a magnitude of 1.1T and points downward through the coil. If the time required for the
field to reverse directions is 0.10 s, what average current flows through the coil during that time?
ε = N (Δϕ /Δt) = IR
= 200 (.022/.1) = I (5Ω)
I = 88 A
t = 0.1s
Bi = 1.1T
Bf = 1.1 T
R = 5Ω
A = 100 cm2
N = 200 turns
Quantitative Problems from Today:
1) Amanda Liewen borrowed from Essentials of College Physics, page 544, #36)
An emf of 24.0 mV is induced in a 500-turn coil when the current is changing at a rate of
10.0 A/s. What is the magnetic flux through each turn of the coil at an instant when the current is
4.00 A?
ε = 24 x 10 -3 V
N = 500 turns
ΔI = 4 A/s
I=4A
ε = -L (ΔI / t) ➜
L = NϕB / I
➜
ε = - (NϕB / I) (ΔI/Δt)
24 x 10 -3 V = - (500 x ϕB / 4A) (10 A/s)
ϕB = -1.92 x 10 -5 Wb
2) Paul Martin borrowed from Midterm 2, Spring 2008 #7
A 20 turn coil of area 10.0 cm2 is placed in a magnetic field so that the normal to its area
is in the direction of the field. If the field originally has a value of 0.25 T that increases
uniformly to 0.35 T in 2.0 s, find the average emf induced in the coil during this time.
** NOTE: Shusaku changed this problem so that Bi = 0.25 Ty and Bf = 0.35 Ty in a magnetic
field where the area was perpendicular to the magnetic field. Remember, calculate θ with respect
to Area in the normal field.
θ = 30°
t = 2.0 s
ϕBf = BA cos θ
Tuesday, July 8th - Notes Prepared by Melanie Smith
N = 20 turns
A = 10cm2 = .001m2 (see below for conversion)
Bi = 0.25 Ty
Bf = 0.35 Ty
ϕBi = BfA cos θ
ε = -N ΔϕB/Δt
ϕBf = (0.35T)(0.001m2)(cos 30) = 3.03 x 10 -4
ϕBi = (0.25T)(0.001m2)(cos 30) = 2.16 x 10-4
ΔϕB = ϕBf - ϕBi = 3.03 x 10 -4 - 2.16 x 10-4 = 8.7 x 10 -5
ε = -20 (8.7 x 10 -5 / 2.0 s) = -8.7 x 10 -4
**Conversion from cm2 to m2
100 cm are in 1 m, so ...
If you have 10cm2 then 10cm2 x (1m/100cm) x (1m/100cm) = 10 x 10-4 m2 or .001 m2