TUTORIAL 6
Problem 1
~ = B0 ~ay W b/m2 where B0 is a constant.
A magnetic flux density is given by B
x
A rigid rectangular loop is situated in the xz-plane with the corners at the point
(x0 , z0 ), (x0 , z0 + b), (x0 + a, z0 + b), (x0 + a, z0 ). If the loop is moving with the
velocity ~v = v0~ax , determine the induced emf.
Proof. At time t the corners of the loop will be at the points (x0 + vt, z0 ), (x0 +
vt, z0 + b), (x0 + a + vt, z0 + b), (x0 + a + vt, z0 ). Using the Faraday’s Law emf =
H
R
~ · d~l = − d
~ ~
~
E
ay implying that
dt S dS · B. In our case dS = dxdz~
C
Z
Z z0 +b Z x0 +a+vt
dx
x0 + a + vt
~
~
dS · B = B0
dz
= B0 b ln
.
x
x0 + vt
S
z0
x0+vt
It follows that
emf = B0 bv
1
1
−
x0 + vt x0 + a + vt
.
Problem 2
Solve the previous problem for a stationary loop in the time-varying magnetic
~ = B0 cos ωt~ay W b/m2 .
field B
x
Proof. If the loop is at rest by analogy with the previous example,
Z
Z z0 +b Z x0 +a
d
dx
~
~
emf = −
dS · B = ωB0 sin ωt
dz
dt S
x
z0
x0
x0 + a
.
= ωbB0 sin ωt ln
x0
Therefore,
emf = ωbB0 sin ωt ln
x0 + a
x0
Problem 3
Assume that the loop in Problem 1 moves with the velocity ~v = v0~ax in the
~ = B0 cos ωt~ay W b/m2 , find the induced emf.
time-varying field B
x
1
2
TUTORIAL 6
Proof. In this case, the loop moves and the magnetic flux density changes with time
such that
Z z0 +b Z x0 +a+vt
d
dx
emf = −B0 cos ωt
dz
dt
x
z0
x0 +vt
d
x0 + a + vt
= −B0 b
cos ωt ln
.
dt
x0 + vt
Evaluating the derivative we find
x0 + a + vt
1
1
emf = B0 b ω sin ωt ln
+ v cos ωt
−
.
x0 + vt
x0 + vt x0 + a + vt
Problem 4
~ =
A rod of length l rotates about the z-axis with the angular velocity ω. If B
B0~az , determine the voltage induced in the rod.
Proof. If we assume the rod was located along the x-axis at t = 0, it follows that
at the time t it makes the angle φ = ωt with the x-axis. Applying Faraday’s law in
the integral form to the sector formed with the x-axis and the position of the rod
at the time t.
Z
Z ωt Z l
d
1
d
1
~ ·B
~ = − d B0
emf = −
dS
dφ
dρρ(~az · ~az ) = − B0 l2 ωt = − B0 l2 ω.
dt
dt
2
dt
2
0
0
And so the voltage induced in the rod is
V = emf =
1
B0 ωl2 .
2
Problem 5
A battery of emf and internal resistance r is hooked up to a variable ”load”
resistance R. If you want to deliver the maximum possible power to the load, what
resistance R should we choose, without changing and r?
Proof. As the maximum current drawn from the system cannot exceed I = r0 (
since V = − Ir0 ) where r0 is the resistance of the system, the maximum current is
I=
r+R
Using Joule’s heating law,
P = I 2R =
2 R
(r + R)2
we take the derivative with respect to R to find the critical point:
1
2R
dP
= 2
−
= 0.
dR
(r + R)2
(r + R)3
Thus to have this vanish we must have r + R = 2R which implies R = r, and so
this is the resistance we should choose to maximize the power to the load.
TUTORIAL 6
3
Problem 6
A square loop is mounted on a vertical shaft and rotateda t angular velocity ω.
~ points to the right. Find the (t) for this alternating
A uniform magnetic field B
current generator.
~ = B0~ay , then the flux is
Proof. Setting B
~ · ~ax = B0 a2 cos θ
Φ=B
where θ is the angle between the x -axis and the unit normal vector to the square
loop. Setting θ = ωt then
dΦ
= −
= −B0 a2 (−sinωt)ω
dt
= B0 ωa2 sin ωt
Problem 7
Determine the mutual inductance between a very long (i.e., infinite) straight
wire and a coplanar rectangular loop with length l and width w, where the distance
between the closest side of the loop and the wire is d.
4
TUTORIAL 6
Proof. To calculate the mutual inductance,v we have to compute the magnetic flux
through the rectangular loop. The magnetic field at a distance r away from the
~ = µ0 I ~aφ by Ampère’s law. The total magnetic flux ΦB through
straight wire is B
2πr
the loop can be obtained by summing over contributions from all differential area
~ = ldr~aφ
elements dA
Z
ΦB =
~ · dA
~ = µ0 Il
B
2π
Z
d
d+w
dr
µ0 Il
=
ln
r
2π
d+w
d
.
Therefore the mutual inductance is,
M=
ΦB
µ0 l
=
ln
I
2π
d+w
d
.
Problem 8
2
A spherical distribution of charge ρv = ρ0 1 − rb2 exists for 0 ≤ r ≤ b. The
charge distribution is concentrically surrounded by a conducting shell with radius
~ everywhere.
ri (≥ b) and outer radius ro . Use Gauss’ law to determine E
Proof. There are four regions to consider 0 ≤ r ≤ b, b < r < ri , ri ≤ r ≤ ro and
~ is of the form E~ar For
r ≥ ro . Due to symmetry, we assume that the vector E
r ≤ b, Gauss’ law implies
Z
~ · dS
~
D
=
Qenc
Z r
r̃2
E(4πr2 ) = ρ0
1 − 2 4πr̃2 dr
b
0
Z r
r̃4
= 4πρ0
r̃2 − 2 dr
b
0
3
5 r
r̃
r̃
= 4πρ0
− 2
2
5b 0
3
r
r5
= 4πρ0
− 2
3
5b
And so, solving for E we see that
~ = ρ0
E
r
r3
− 2
3 5b
~ar
TUTORIAL 6
5
For b < r < ri , Gauss’ law implies
Z
~ · dS
~ =
D
Z
Qenc = ρv dV
Z b
r̃2
0 E(4πr2 ) = ρ0
1 − 2 4πr̃2 dr
b
0
Z b
r̃4
2
= 4πρ0
r̃ − 2 dr
b
0
3
b
r̃
r̃5
= 4πρ0
− 2
2
5b 0
3
b
b5
= 4πρ0
− 2
3
5b
And so, solving for E we see that
3
b3
2ρ0 b3
b
~ = ρ0
E
−
~ar =
~ar
2
0 r
3
5
150 r2
~ = 0.
For ri ≤ r ≤ ro , the conducting shell causes E
For r > ro , Gauss’ law:
Z
0
~ · dS
~ = Qenc =
E
Z
ρv dV
implies this will be the same as b < r < ri :
3
~ = 2ρ0 b ~ar
E
150 r2