NAME:
ELECTRICITY AND MAGNETISM II
First Midterm Exam (Spring 2016)
Show all your work to receive full credit
#
1 : (15 points)
Two coils are wrapped around a cylindrical form in such a way that the same flux passes
through every turn of both coils. The primary coil has has N 1 turns and the secondary
has N 2 turns (see figure). This ideal transformer takes an input AC voltage of amplitude
V1 , and delivers an output voltage of amplitude Vi, which is determined by the turns ratio
Secondary
(N2 turns)
Primary
(N 1 turns)
(a) In an ideal transformer, the same flux passes through all turns of the primary and of
the secondary. Show that in this case M 2 = L 1 L 2 , where Mis the mutual inductance
of the coils, and L 1 , L 2 are their individual self-inductances. (3 pts.)
(b) Suppose the primary is driven with AC voltage
Vin
=
Vi cos(wt),
and the secondary
is connected to a resistor, R. Show that the two currents satisfy the relations
dli
L1 dt
+M
dh
dt =Vi cos(wt)
L 2 dh
dt
+ Mdli
dt
=-I2 R
.
(3pts. )
(c) Using the result in (b), solve these equations for Ii (t) and I 2 ( t). (Assume Ii has no
DC component.) (3 pts.)
(d) Show that the output voltage (Vout = I2R) divided by the input voltage CVin is equal
to the turns ratio: Vout/Vin = N2/ Ni. (3 pts.)
(e) Calculate the input power (Pin = Vinli) and the output power (Pout = Vouth), and
show that their averages over a full cycle are equal. (3 pts.)
#
2 : (10 points)
An infinitely long cylindrical tube, of radius a, moves at constant speed v along its axis. It
carries a net charge per unit length .A, uniformly distributed over its surface. Surrounding
it, at radius b, is another cylinder, moving with the same velocity but carrying the opposite
charge (-..\).
(a) Find the electric and magnetic fields between the cylinders. (2 pts.)
(b) Find the energy per unit length stored in the fields. (3 pts.)
(c) Find the momentum per unit length in the fields. (3 pts.)
(d) Find the energy per unit time transported by the fields across a plane perpendicular
to the cylinders. (2 pts.)
Hint: g = to(E
x B)
Srkh'ffrl ~ f~ni
11=I1L1+MI2 :M~
(().)
X_
:.e
~
I L1 + I 11
j
'M
;l
·r
A}, = _._.)_
#i
, f;-=~(,_+i1IJ:~p
L2 . T H
N2 ·+ _, N2
fv1 _ N,_ _ L2
-------L,
N,
M
(b)
{c)
-E,
=
- Ei
::::
~1
=
L,*
+11 ~
clf-i. ~ L:i. JT:z. + 11 dI1
di
df
di
=
~eei(c.it)
R
== - I
'2
mi.JJ;f11 ~~11. bd '-2
JI + L MiJ_
L, L:i -;;rf
Jl,_ = L2 v; ®(c.it)
2
Qmj 1~.ut L2 ~· .~ .woid. ~11.
~ + H(-1).. 'R -JJ1ti)"' L/v', f.M(l!t)
"
Stl1Cl..
L 1L2
=ML- .
1
l-leViO-
I2 ft) -:::: - L..z V, un(lJ ·t)
M~
N'ffW ,bi StK+
Tl. ,f)
,~
-fl r·yif 1, , ...
L.,Jr, + M Li.V, fA)s1nftJ-t) ~V,U?AfLJt)
di
M~
rJr,
== _
LI I?.
J.j
~:_
~-
Lz ~ w siYJ (tJf)
L2 Vt un(wt) +
Lj R
+Ji 4nfwt)
L,
~l
L,
w
Sih{wt)::::;
----------.
ll_
L,
[J_ srn{wt) + blM/wr)l'j
CJ
1<.
r,J.&wi :tic 1 ~
(J)
~:::: T 2
V,·H
R __
v; ~{wt)