Download HW # 3

5.12, 5.13, 5.16, 5.20, 6.1, 6.8, 6.11, 6.15 - Due Oct. 14, 2011
Problem 5.12:
Two infinitely long, parallel wires 2m apart are carrying 6-A currents in opposite directions.
Determine the magnetic flux density at point P, 0.5m from the first wire.
Equation 5.30 states
̂
, thus
̂
̂
(into the
page).
Problem 5.13:
A long East-west oriented power cable carrying an unknown current I is at a height of 8m above
the Earth’s surface. If the magnetic flux density recorded by a magnetic field meter placed at the
surface is 15μT when the current is flowing through the cable and 20μT when the current is zero,
what is the magnitude of I?
As no direction is stated, it is sufficient to simply give the magnitude of the current. Since the
difference between the magnetic flux density is 5μT, we have
, from which we get
I = 200A
Problem 5.16:
A long, straight conductor lies in the plane of a rectangular loop at a distance d = 0.1m to the left.
The loop is 0.2m wide and 0.5m tall. The long straight conductor has a current of 20A, in the
upward direction, and the loop has current of 20A in the clockwise direction. Determine the net
magnetic force acting on the loop.
We can ignore the effects of the magnetic field due to the current in the segments of the loop that
run perpendicular to the current I1; since the conductor is long, the two rotational forces cancel
each other out. We can not ignore the fact that it connects the two currents that run parallel to I1,
ultimately meaning that we can just add the two forces together to get the total.
If we adopt the coordinate system of Figure 5-14, we can use equation 5.40:
̂
̂
0.4mN (towards I1)
Problem 5.20:
A square loop placed at the origin in the positive xy plane has 2m sides and carries a current of 5A
in the clockwise direction. If a straight long conductor carrying a current of 10A is introduced and
placed just above the midpoints of two of the loops sides, determine the net force acting on the
loop.
We can ignore the current running perpendicular to the long wire, i.e.:
Take the current in segment 2 for the portion in the
⁄ area. Current is in the
̂ direction,
and the magnetic field can be broken down into only ̂ and ̂ components. The y components
vanish, since they are parallel to the direction of current, however the z components will lead to
an overall force in both the –x and +y directions. Similarly, the part of segment 2 in the
⁄
area an overall force will occur in the –y and +x directions. The y components will cancel, leaving
the net force in the y direction 0. However, because the x components are equal and opposite
about an axis point, they will lead to a rotational force. Fortunately, the current in segment 4
yields an equal and opposite rotational force. Much like the previous problem, the sum of forces
due to the currents in the loop that run perpendicular to I2 is equal to zero.
So, we have
̂
⁄
̂
⁄
(to the right).
Problem 6.1:
The switch in the bottom loop of figure P6.1 is closed at t=0 and then opened at a later time t1.
What is the direction of the current I in the top loop at these two times?
At t=0 we have a change in current in the clockwise direction. This change in current results in a
change in magnetic flux. The direction of the change in flux – not necessarily the direction of the
flux itself - is into the page at the bottom loop and out of the page at the top loop. Lenz’s law
states that the current in the top loop will be in a direction that opposes this change – the
direction of current that would cause a change in flux into the page at the top loop is clockwise at
t=0.
At t = t1, the direction of the change in flux is out of the page at the bottom loop, and into the page
at the top loop. So, for the top loop, we are looking for the current that would cause a change in
flux out of the page, which is counter-clockwise at t=t1.
Problem 6.8:
The transformer shown in figure P6.8 consists of a long wire coincident with the z-axis carrying a
current I = I0 Cos ωt, coupling magnetic energy to a torodial coil situated in the x-y plane and
centered at the origin. The torodial core uses iron material with relative permeability μ1. Around
which 100 turns of tightly wound coil serves to induce a voltage Vemf, as shown in the figure.
a) Develop an expression for Vemf.
b) Calculate Vemf for f = 60Hz, μr = 4000, a = 5cm, b = 6cm, c = 2cm, and Io = 50A.
a)
. The magnetic flux is defined as
∫
. Equation 5.30 gives us
, and our integration will be from a to b, making
. Since we can see by
inspection that both B and ds are in the same direction, we can put these into the
equation directly, giving
⁄
b)
∫
( ⁄ )
. Thus,
Problem 6.11:
The loop shown in P6.11 moves away from a wave carrying a current I1 = 10A at a constant velocity
u = y7.5 m/s. If R = 10 ohms and the direction of I2 is as defined in the figure, find I2 as a function of
y0, the distance between the wire and the loop. Ignore the internal resistance of the loop.
Knowing
,
̂
∫
, and
rectangle to find
. We can integrate over the surface of the
, where y0 is a function of t.
Taking the negative derivative of this with respect to time, we get
[
], where
is the change in position with respect to the
change in time – more commonly known as the velocity, which is given as 7.5.
[
And
]
[
]
.
The direction of the current can be seen from inspection. We know the direction of the change in
magnetic flux is out of the paper, as the magnetic field weakens as y0 increases. The current
required to generate a flux into the paper, thus opposing this change, is in the clockwise direction.
Problem 6.15:
A coaxial capacitor of length l = 6cm uses an insulating dielectric material with εr = 9. The radii of
the cylindrical conductors are 0.5cm and 1cm. If the voltage applied across the capacitor is V(t) =
50sin(120πt) what is the displacement of the current?
We know
, and we are given V. C can be calculated by equation 4.116,
.
. So,
⁄