Download EXAM 3: SOLUTIONS Q1.The normal to a certain 1m area makes an

EXAM 3: SOLUTIONS
Q1.The normal to a certain 1 m2 area makes an angle of 60o with a uniform
magnetic field. The magnetic flux through this area is the same as the flux
through a second area that is perpendicular to the field if the second area is:
SOLUTION
(1)
~ A
~ 1 = BA1 cos 60o
ΦB = B.
(2)
~ A
~ 2 = BA2 cos 0o = BA2
ΦB = B.
A1 = 1 m2
(1)
(2)
A2 =?
ΦB = ΦB ⇒ A2 = A1 cos 60o =
A1
= 0.5 m2
2
Q2.A rectangular loop of wire is placed midway between two long straight
parallel conductors as shown. The conductors carry currents i1 and i2 as
indicated. If i1 is increasing and i2 is constant, then the induced current in
the loop is :
SOLUTION
The currents i1 produces a magnetic field which is directed into the page and
the current i2 produces a magnetic field which is directed out of the page.
The total flux in the loop has contributions from both the fields but only the
former is incresing with time while the latter is a constant. The total flux
depends on the relative strengths of i1 and i2 . There are two cases, i1 (t) > i2
and i1 (t) < i2 . For both of them Lenz’s law gives the direction of induced
current as counter-clockwise.
i1 (t) > i2
i1 (t) < i2
Φ̂net
⊗-increasing
-decreasing
Φ̂ind
îind
CC
CC
(NOTE: CC - Counter-clockwise.)
1
Q3.The circuit shown is in a uniform magnetic field that is into the page.
The current in the circuit is 0.20 A. At what rate is the magnitude of the magnetic field changing? Is it increasing or decreasing?
SOLUTION
Vnet = Vapp + ξind = iR.
ξind = iR − Vapp .
Now, the direction of current was not mentioned. This ambiguity is resolved
by choosing a sign convention for the current i. If we choose i to be positive
for clockwise direction and negative otherwise, then
(
ξ = iR − Vapp =
(0.2 × 10) − 4 = −2 V
(−0.2 × 10) − 4 = −6 V
if î is clockwise
if î is counter-clockwise
The opposing emf induces a flux which is directed into the page indicating
that the primary flux is decreasing.
To find the rate of change of the primary field, use the Faraday’s law
ξ=−
ξ
dB
=− =
dt
A
(
dΦB
dB
= −A
dt
dt
(2/(12 × 10−2 )2 ) = 140 T/s
(6/(12 × 10−2 )2 ) = 420 T/s
if î is clockwise
if î is counter-clockwise
If we relax the sign condition on current then we have two correct choices
and both are considered right answers.
Q4We desire to make an LC circuit that oscillates at 100 Hz using an inductance of 2.5 H We also need a capacitance of :
SOLUTION
f = 100 Hz; L = 2.5 H; C =?
1
1
ω = 2πf = √
⇒ C = 2 2 = 1.01 × 10−6 F ' 1µ F.
4π f L
LC
2
Q5.An LC circuit has a capacitance of 30µ F and an inductance of 15 m H.
At time t = 0 the charge on the capacitor is 10µ C and the current is 20 mA.
The maximum charge on the capacitor is:
SOLUTION
The charge and current through the circuit at time t is
i(t) = −Qω sin(ωt + φ)
q(t) = Q cos(ωt + φ)
Defining q0 ≡ q(t = 0) and i0 ≡ i(t = 0),
q0 = Q cos φ ⇒ cos2 φ =
q02
Q2
i0 = −Qω sin φ
i20 = Q2 ω 2 sin2 φ = Q2 ω 2 (1 − cos2 φ)
!
q02
i20
2
= Q 1 − 2 = Q2 − q02
2
ω
Q
s
Q =
q0 = 10µ C;
i0 = 20 mA;
q02 +
q
i20
=
q02 + i20 LC
ω2
L = 15 mH;
C = 30µ F.
Q = 16.73 × 10−6 C ' 17µ C.
Q6.The primary of a 3 : 1 step-up transformer is connected to a source and
the secondary is connected to a resistor R. The power dissipated by R in
this situation is P . If R is connected directly to the source it will dissipate a
power of :
SOLUTION : For a step-up transformer, Ns > Np
Ns
Vs
=3= ;
Np
Vp
Rpeq = R
Np
Ns
2
= R/9
With the resistor connected in the secondary circuit, power is
P =
Vp2
9Vp2
=
Rpeq
R
If the same resistor is connected in the primary circuit, power is
P0 =
Vp2
P
=
R
9
3
Q7.A bar magnet is placed vertically with the S pole up and the N pole down.
~ field at its center is :
The B
SOLUTION
The field lines emerge out of the N pole and enter into the S pole. Inside
the magnet they go S-N. So with the S pole facing up, the direction of the
~ field will be down.
B
Q8.The impedence of the circuit shown is : (Ref. to the picture)
SOLUTION :
ω = 2πf = 200π rad − s−1 ;
L = 0.5 H;
C = 200µ F;
R = 100Ω.
1
50
= Ω.
Cω
π
q
√
Impedence, Z = R2 + (XL − XC )2 = 3 × 102 = 173Ω.
XL = Lω = 50πΩ
XC =
Q9.The diagram shows one plate of a parallel-plate capacitor from within the
capacitor. The plate is circular and has a radius R. The dashed circles are
four integration paths that have radii of r1 = R/4, r2 = R/2, r3 = 3R/2 and
H
~ s around the
r4 = 2R. Rank the paths according to the magnitude of B.d~
paths during the discharging of the capacitor, least to greatest.
SOLUTION
Between the plates of a discharging capacitor, the electric flux is changing
with time giving rise to a fictitious displacement current id . This current
produces a circular magnetic field around the plate’s central axis. By the
H
~ s = µ0 id (r). If id is the net displacement current
Maxwell-Ampere’s law, B.d~
between the plates, then the displacement current density is Jd = id /(πR2 ).
The displacement current enclosed by a circular path of radius r is id (r),
where


2
J × πr 2 = id
id (r) =  d
Jd × πR2 = id
I


µ0 id
~ s =
B.d~
 µ i
o d
2
r
R
4
r
R
(interior)
(exterior)
(interior; r1 , r2 )
(exterior; r3 , r4 )
I
2
~ s = µ0 id r1 = µ0 id
B.d~
R
16
1
2
I
I
r
µ
~ s = µ0 id 2 = 0 id = 4 B.d~
~ s
B.d~
R
4
2
1
I
I
I
~ s =
~ s = µ0 id = 16 B.d~
~ s
B.d~
B.d~
3
4
I
1
~ s<
B.d~
I
2
1
~ s<
B.d~
5
I
3
~ s=
B.d~
I
4
~ s.
B.d~