TEST 4
(PHY 250)
1. a)
b)
c)
d)
What is the magnetic hysteresis?
Define the displacement current.
Write the Lenz rule.
Write the definition of the magnetic field vector.
2. a) Using the Amprere’s law, relate the magnetic field Bin inside a uniformly
wound solenoid having N turns and length L a solenoid to the current in the
solenoid. Assume that the diameter of the solenoid is much smaller than its
length
b) From the obtained relationship find its inductance if a ferromagnetic core is
inserted into the solenoid.
3. In the attached figure the current in the long straight wire is I1 = 5A, and the wire lies
in the plane of the rectangular loop, which carries 10A current. The marked
dimensions are c = 0.1 m, a = 0.15m, and l = 0.45m.
a) Determine1 the magnetic force exerted by the wire on the loop.
b) Find the magnetic moment of the loop and the magnetic torque exerted on the
loop by the wire.
4. A rectangular loop of width a and length l is located near a long wire carrying
current I1, as in the previous problem. Derive an expression for the flux of the
magnetic field produced by the wire over the surface spanned on the loop.
5. Consider the current-carrying loop shown in the attached figure, formed of radial
lines and segments of circles whose centers are at point P. Find the magnetic field at
point P.
1
µ0 = 4π⋅10-7 Tm/A
-1-
a) In ferromagnetic substance the magnetization is sustained after the
external magnetic field is removed. This effect is called magnetic hysteresis.
b) The displacement current (through a surface) is associated with the
change in the electric flux though the considered surface
I d = ε0
dΦ E
dt
c)
The polarity of an induced electromotive force is such that it tends to
produce an induced current creating a magnetic field that opposes the
change in the flux causing the electromotive force.
d) The magnetic field vector B(r) at position r is a vector such that, at this
position, the magnetic force exerted on a particle with charge q, moving with
velocity v, would be
FB = qv × B
-2L
a) We can assume that in the solenoid the
magnetic field is uniform directed along the
solenoid’s axis, and outside in the solenoid
the magnetic field has a zero value. (The
assumptions are based on the symmetry of I
the solenoid.) The linear integral of the
magnetic field along in amperian loop
indicated in the figure (the circulation of the magnetic field) is therefore
∫ Bds = BL
The total current flowing through a surface spanned by the contour depends
on the current in the solenoid’s wire and the number of loops
I tot = NI
Amper’s law relates these two quantities
BL = µ 0 NI
from which
N
B = µ0 I
L
b) We have to find now the proportionality coefficient between the current in
the solenoid and the voltage across the solenoid (the electromotive force
induced in the solenoid). From the above equation we can relate the
magnetic flux in the solenoid with a core to the electric current
NA
Φ B = B' A = (1 + χ)µ 0
I
L
where B’ is the magnetic field in the solenoid with the core, A is the
cross-sectional area of the solenoid and χ is the susceptibility of the core
material. According to Faraday’s law of induction, the electromotive force is
related to the rate of change in the magnetic flux, which can be related from
the above equation to the current
dΦ B
N 2 A dI
V = −N
= −(1 + χ )µ 0
dt
L dt
Hence the inductance of the solenoid is
N2 A
L = (1 + χ )µ 0
L
B
-3y
B
I1
I2
c
µ
a
a
b
l
a) Using the right-hand rule we can find the
magnetic field of the wire in perpendicular to the
plane of the loop. By symmetry, we note that the
magnetic forces on the top and the bottom
segments of the rectangle cancel. From the
Amper’s law, the magnitude of the magnetic field
(produced by the wire) at distance c from the wire
has a value
x
B=
µ 0 I1
2 πc
(or its scalar components in the indicated coordinate system are
µ I 

B =  0,0, − 0 1  )
2 πc 

From the definition of the magnetic field vector the magnitude of force
exerted on the closer side of the loop is therefore
Fa = BI2l =
µ0 I1I 2l
2 πc
with the direction "to the left”
(or in the vector form
 µ II l

Fa =  − 0 1 2 ,0,0  )
 2πc

Similarly, the magnitude of force exerted on the farther side of the loop is
Fb = BI 2l =
µ 0 I1I 2l
2π(c + a )
with the direction “to the right”
(or in the vector form
 µ IIl

Fb =  0 1 2 ,0,0  )
 2π(c + a )

So the force exerted on the loop is
F=
=
µ 0 I1I 2l  1
1
−  î =

2π  (c + a ) c 
1 
Tm 
 1 − 1  î = −27µNî
⋅  4 π ⋅ 10− 7
⋅
5
A
⋅
10
A
⋅
0
.
45
m
⋅



2π 
A 
 0 .25m 0 .1m 
b) Using the right-hand rule we can find that the direction of the magnetic
moment coincides with the direction of the magnetic field produced by the
wire (as I marked in the figure). The magnitude on the magnetic moment can
be found from the area of the loop and the value of the current in the loop
µ = I 2 al = 10 A ⋅ 0.15m ⋅ 0.45m = 0.675Am2
Since, the magnetic moment and the magnetic field are parallel the magnetic
torque is a zero vector.
-4y
B
dy
I1
I2
dx
b
a
l
The differential flux over a fragment separated
from the wire by distance x depends on the size,
the relative orientation and the value of the
magnetic field
dΦ B = B(x , y ) ⋅ dxdy ⋅ cos 0°
As indicated in the above equation, the magnetic
field is a function of the segment. Before we can
c
a
integrate the flux, we need to express the
x
magnetic field explicitly in term of the variables of integration (x and y are a
convenient choice)
B(x , y ) =
µ 0I1
2 πx
The magnetic flux is therefore
l c+ a µ
ΦB = ∫ ∫
0 c
0 I1
2 πx
µ0 I1 c + a
µ Il c+a
ln
dy = 0 1 ln
c
2π
c
0 2π
l
dxdy = ∫
-5From the Biot-Savart law, the
differential contribution to the magnetic field
at point P is
(ii)
(iii)
I ds b
b
dB =
r
(iv)
I
α
(i)
(v)
I dsa
n
µ0 Ids × r$
⋅ 2
4π
r
In order to find the magnetic field
vector at this point we have to "add" all
differential fields.
P
a
B=
µ 0 Ids × rˆ
∫ 4π ⋅ r 2
wire
Practically, the rest is math. We will try to choose such a variable and
to divide the wire into such pieces so that the integration is easy. Notice that
for the straight fragments the direction of the elements ds coincides with the
direction of the unit vector r$ from the segment to the considered point P.
The vector product of these two vectors is a zero vector. Directly from the
definition of the integral, we can conclude that the three straight fragments
of the wire do not contribute to the magnetic field at point†P.
Let's choose angle α as the variable of integration for the two curved
fragments. In this case
Idsb × r$ = − Ibdα ⋅ n$ and Idsa × r$ = Iadα ⋅ n$
where n$ is the unit vector directed "out of the page". In summary
π
3
π
3 µ
 µ Ibnˆ dα 
 0 Ianˆ dα 
B = ∫ dB + ∫ dB + ∫ dB + ∫ dB + ∫ dB = 0 + ∫  − 0 ⋅
+
0
+

 ⋅
 +0 =
∫
2
b 
a2 
i
ii
iii
iv
v
0  4π
0  4π
µ Ibnˆ α
=− 0⋅ 2
4π b
π
3
0
π
µ Ianˆ α 3 µ 0 I  1 1 
+ 0⋅ 2
=
⋅  −  ⋅ nˆ
4π a 0 12  a b 