Download Physics 232 : Midterm 3 Practice 1 Problem 1 (25 points)

Physics 232 : Midterm 3 Practice
November 17, 2016
Please show all the details of your computations including intermediate steps.
1
Problem 1 (25 points)
Consider the left and middle figures.
a) Compute the magnetic field at point P distance r = a away from a thin, straight wire of finite length
~ the magnetic field
carrying a constant current I and placed along the x axis. Do this by setting up dB
~ by changing the
due to the small line segment d~s in the middle figure. Then directly integrate dB
variable x = −a tan θ. The Biot-Savart Law for magnetic field is given in the formula sheet.
Sol : Biot-Savart Law is given by
Z
µ0 I
d~s × ~r
~ = µ0 I
B
=
(sin θ1 − sin θ2 )ẑ .
4π
r3
4πa
Here this quantity is always positive due to our convention, θ1 > 0, θ2 < 0.
~ as
The computation is following. Using the following figure we compute dB
d~s = dxx̂,
~r = aŷ − xx̂,
d~s × ~r = adxẑ ,
~ =
dB
µ0
Iadx
µ0 I
ẑ = −
cos θdθẑ ,
4π (a2 + x2 )3/2
4πa
where we use x = −a tan θ (thus x < 0 corresponds to θ > 0).
b) Using the result a), compute the infinite line of current a distance a away. Compare this result with
that you can compute using the Ampere’s law.
Sol : Infinite line corresponds to θ1 = −θ2 = π/2
~ = µ0 I ẑ .
B
2πa
This is consistent with the result from Ampere’s law.
Consider the right figure.
c) Using the result obtained in a), compute the magnetic field at point P , located a distance x from
the corner of the infinitely long wire carrying a current I, which is bent at a right angle.
Sol : The current is similar to the case θ1 = 0 and θ2 = −π/2. Thus (into the page)
~ = − µ0 I ẑ .
B
4πx
2
Problem 2 (25 points)
A rectangular conducting loop of width a and length b is located near a long wire carrying a current
I~ = (p + qt)ŷ with p, q constant and t time. The distance between the wire and the closest side of
the loop is c. The wire is parallel to the long side of the loop initially and rotates with an angular
velocity ω along the axis running through the center located at x = c + a/2. Thus the angle between
the magnetic field and the ‘direction’ of the loop is θ = 180 + ωt for our convention.
Note that we attach our coordinate system such that x, y, z are to the right, up along the current
direction and out of the page.
a) Compute the emf generated on the loop. Use all the explicit time dependences.
~ = − µ0 I(t) ẑ = − µ0 (p+qt) ẑ Using the flux passing through the loop
Sol : Using Ampere law, we have B
2πr
2πr
can be computed with the angle
Z
Z b Z c+a
d
d
d µ0 (p + qt)
c+a
µ0 (p + qt)
~ · dA
~= d
B
dy
cos(ωt)
dr
E =− Φ=−
cos(ωt) =
b ln
dt
dt
dt 0
2πr
dt
2π
c
c
c+a
µ0
b ln
=
[q cos(ωt) − ω(p + qt) sin(ωt)] .
2π
c
where − sign in the first line cancels the relative directions between the magnetic field and initial loop.
b) Compute the current running through the loop.
Sol : Using E = IR, we get
µ0
c+a
I=
b ln
[q cos(ωt) − ω(p + qt) sin(ωt)] .
2πR
c
3
Problem 3. (25 points)
R
Consider the left figure.
a) Consider a cross section of an ideal solenoid with N turns of current I carrying wire per length `
and radius R. Use the Ampere’s law, compute the magnetic fields everywhere. Specify the Ampere
loops you use to compute the magnetic field. Use the coordinate system such that positive x, y, z axes
are towards up, out of the page and to the right.
Sol : We did this in class. Using the loop 1, we can show that the magnetic field outside the solenoid
vanishes. Using the loop 2, the Ampere law gives us `B = µ0 N I, thus B = µ0 I N` ẑ inside the solenoid.
B = 0 outside.
b) The current is given by I(t) = I0 cos(ωt) for t > 0. Compute the emf generated for t > 0.
~ ·A
~ = µ0 N I0 cos(ωt)πR2 /` for a single
Sol : Faraday’s law tells that the emf is given by (using ΦB = B
loop)
E = −N
dΦB
µ0 N 2 A dI(t)
µ0 N 2 I0 πR2 d cos(ωt)
µ0 N 2 I0 πωR2
=−
=−
=
sin(ωt) .
dt
`
dt
`
dt
`
~ due to the emf generated in b).
c) OPTIONAL : Compute the electric field E
Sol : Faraday’s law of induction and the relation to electric field is given by
I
dΦB
~ · d~s .
E =−
= E
dt
The change of the magnetic field is along z-direction. To create this we need electric field in the same
direction as the current direction. The integration of the electric field is similar to the Ampere law
computation for loop 1.
H
~ · d~s = − dΦB gives
Thus for r > R, the equation E
dt
2
µ0 N I0 πωR2
~ = µ0 N I0 ωR sin(ωt)θ̂ .
sin(ωt),
E
`
2` r
H
~ · d~s = − dΦB gives (the area is changed from πR2 to πr2 )
For r > R, the same equation E
dt
2πrE(r) =
2πrE(r) =
µ0 N I0 πωr2
sin(ωt),
`
~ = µ0 N I0 ωr sin(ωt)θ̂ .
E
2`
Consider the right figure. For simplicity, assume that the magnetic field inside the finite solenoid
is the same as the infinite-long ideal solenoid.
d) Compute the emf generated in the coil 2 with radius R2 (NH turns with length `2 ) when there are
currents I1 (t) in the coil 1 with radius R1 (NB turns with length `1 ) and I2 (t) in the coil 2. What are
the self and mutual inductances?
Sol : The emf contains the contributions from the magnetic flux due to the currents I1 (t) and I2 (t),
which have the information on the self and mutual inductances between two circuits with
2
dΦB,22
dΦB,21
µ0 NH
A2 dI2 (t) µ0 NH NB A1 dI1 (t)
− NH
=−
−
dt
dt
`2
dt
`1
dt
dI2
dI1
= −L2
− M12
.
dt
dt
EL (2) = −NH
Where A2 = πR22 , A1 = πR12 . In the second contribution, the area is the area of the solenoid of small
radius because the magnetic field only exist there. Thus we can figure out L2 and M
L2 =
2
µ0 NH
A2
,
`2
M=
µ0 NH NB A1
.
`1
4
Problem 4 (25 points)
Consider the circuit shown in the Figure.
a) Obtain the equivalent inductance Leq of the inductors with inductance L1 , L2 and mutual inductance
M = M12 = M21 and the equivalent resistance Req of the two resistors R1 and R2 . You can use the
induced emf of each inductor and the fact that the currents passing through these inductors add up.
dI2
dI2
dI1
1
Sol : We use the inductance relations E1 = −L1 dI
dt −M dt , E2 = −L2 dt −M dt . We know E = E2 = E1
and I = I1 + I2 .
Now by equating these two relations as E2 = E1 , we obtain
M − L1 dI1
dI2
=
.
dt
M − L2 dt
By using I = I1 + I2 , we get
dI
dI1
dI2
dI1
M − L1 dI1
2M − L1 − L2 dI1
=
+
=
+
=
.
dt
dt
dt
dt
M − L2 dt
M − L2
dt
Using these two equations, we get
E = −Leq
dI2
M − L1 dI1
dI
dI1
dI1
= −L1
+M
= −L1
+M
dt
dt
dt
dt
M − L2 dt
2
M − L1 L2 dI1
M 2 − L1 L2
,
Leq =
=−
2M − L1 − L2 dt
2M − L1 − L2
where we use the first line. On the other hand the equivalent resistance is Req = R1 + R2 for series
resistors.
b) At t < 0, the switch was closed to a. After long time, what is the current passing through the
battery?
E
Sol : I0 = R+R
.
eq
c) At t = 0, the switch S is thrown to b. Obtain the loop equation for the current I(t) of the circuit
using Kirchhoff’s loop rule. Solve the equation using the initial condition given in b).
Sol: With the current direction to be clockwise and loop direction counterclockwise, we get
I(t)Req + Leq
dI(t)
=0,
dt
R
eq
− Leq
t
I(t) = I0 e
d) Compute the time constant for the system.
Sol : The time constant of the system is
L L −M 2
1 2
Leq
τ=
= L1 +L2 −2M .
Req
R1 + R2
.
Physics 232 : Formula sheet for Midterm 3
1. Coulomb’s law :
q1 q2
~ 12 ,
F~12 = ke 2 r̂12 = q2 E
r12
where F~ , ke , q1 , q2 , ~r12 are the force, ke =
2
with the 0 = 8.85 × 10−12 NCm2 as the permittivity of
~ 12 is the electric field due
the free space, charges q1 and q2 , and the vector from the charge q1 to q2 . E
to the charge q1 to the position of the charge q2 .
1
4π0
~ :
2. Electric field E
~ = ke
E
Z
dq
r̂ ,
r2
is the electric field due to the continuous charge distribution. dq can be dq = λdr for the line charge
density λ, dq = σdA for the area charge density σ, dq = ρdV for the volume charge density ρ.
3. Gauss’s law :
I
Φ=
~ · dA
~ = Qin ,
E
0
~ Qin are the electric flux, area vector (with direction outward normal to the surface), and
where Φ, dA,
the net charge enclosed in the Gaussian surface.
~
4. Electric Potential V and field E:
Z
B
V (B) − V (A) = −
A
~ · d~s ,
E
~ = − ∂V î − ∂V ĵ − ∂V k̂ ,
E
∂x
∂y
∂z
where A, B are two different locations that are connected by a path through the infinitesimal vector ~s.
Electric potential V can be also directly evaluated for the continuous charge distribution
Z
dq
V = ke
.
r
~ are the infinitesimal displacement and area element.
5. Coordinate systems, d~r and dA
A. Spherical coordinate : (r, θ, φ)
d~r = drr̂ + rdθθ̂ + r sin θdφφ̂ ,
~ = r2 sin θdθr̂ + r sin θdrdφθ̂ + rdrdθφ̂
dA
B. Cylindrical coordinate : (r, θ, z)
d~r = drr̂ + rdθθ̂ + dz ẑ ,
~ = rdθdzr̂ + drdz θ̂ + rdrdθẑ
dA
6. A few useful integrals :
Z
h
i
p
Log y 2 + y x2 + y 2
dx
Log[x]
−
,
y
y
x
+
Z
Z
p
p
dx
xdx
2
2
p
p
= Log[x + x + y ] ,
= x2 + y 2 .
x2 + y 2
x2 + y 2
p
x2
y2
=
7. Capacitance :
Q = C∆V ,
1
Q2
1
Q∆V =
= C(∆V )2 ,
2
2C
2
C(dielectric) = κC0 .
U=
where C is capacitance, and U is energy stored, κ is the dielectric constant.
8. Electric Dipole :
p~ = 2aqr̂
~
τ = p~ × E
~
U = −~
p·E
Electric dipole moment ,
torque ,
energy .
where 2a is the distance between the positive and negative charges, r̂ is the unit vector from the
negative charge to positive charge.
9. Current & Resistance :
I=
J~ =
dQ
dt
I~
current ,
current density ,
Area
∆V
Length
R=
=ρ
Resistance ,
I
Area
∆V 2
dU
= I∆V = I 2 R =
Power .
P =
dt
R
10. DC circuit & Kirchhoff rules:
E = IR
emf ,
X
Ii = 0
Junction rule : sume of current should vanish at each junction ,
i
X
a
∆Va = 0
Loop rule : sume of potential drops should vanish for each loop .
11. Magnetic force :
~ or
The magnetic force depends on the velocity ~v of the particle with charge q and the magnetic field B,
alternatively if a straight conductor of length L carries a current I, the force exerted on that conductor
~ is
when it is placed in a uniform magnetic field B
~ = IL
~ ×B
~ .
F~ = q~v × B
The unit of magnetic field is T = N/A · m.
~ placed in magnetic field is given by
The torque and energy on the magnetic dipole moment µ = I A
~ ,
~τ = µ
~ ×B
~ ,
UB = −µ · B
µ0 = 4π × 10−7 T · m/A .
12. Biot-Savart Law :
~ = µ0 I
B
4π
d~s × r̂
.
r2
Z
Ampere’s law and the definition of magnetic flux
I
~ · d~s = µ0 Ien ,
B
Z
ΦB =
~ · dA
~.
B
13. Faraday’s law of induction and the relation to electric field :
I
dΦB
~ · d~s .
E =−
= E
dt
Lenz’s law : The induced current in a loop is in the direction that creates a magnetic field that opposes
the change in magnetic flux through the area enclosed by the loop.
14. Inductance :
EL = −
dΦB
dI
≡ −L
.
dt
dt
The emf contains the contributions from the self and mutual inductances between two circuits with
currents I1 (t) and I2 (t)
EL (1) = −L1
dI2
dI1
− M21
dt
dt
EL (2) = −L2
dI2
dI1
− M12
.
dt
dt
M12 = M21 .
15. Solution for a second order differential equation :
ẍ = −ω 2 x,
x = A cos(ωt + φ) .