ELECTRICITY AND MAGNETISM Tutorial (PHYS2239)
BSc, Physics - PHYS
Semester: Second Semester
Level: 200
Year: 2018
Tutorial sheet on Magnetism
1. A cylindrical wire of permeability ji carries a steady current I. If the radius of the wire is R. find B and 11
inside and outside the wire.
Solution:
Use cylindrical coordinates with the z-axis along the axis of the wire and the positive direction along the current
flow, as shown in Figure 1. On account of the uniformity of the current the current density is
J=
Figure 1:
Consider a point at distance r from the axis of the wire. Ampere’s circuital law,
where L is a circle
of radius r with centre on the z-axis, gives for r>R,
H(r)=
or
B(r)=
Since by symmetry H(r) and B(r) are independent of For r < R.
I(r) =
=
And the circuital law gives
H(r)=
, B(r)=
2. A long non-magnetic cylindrical conductor with inner radius a and outer radius b carries a current I. The
current density in the conductor is uniform. Find the magnetic field set up by this current as a function of radius
(a) inside the hollow space (r < a):
(b) within the conductoi (a < r < 6);
(c) outside the conductor (r > 6).
Solution: Use cylindrical coordinates as in Problem 1. The current density in the conductor is
The current passing through a cross-section enclosed by a circle of radius r, where a. < r < b, is
I(r)=
=
!"
#
!"
By symmetry, Ampere’s circuital law gives
(a) B=0, (r>a)
(b) B(r) =
.
!"
#
!"
, (a<r<b)
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(c) B(r) =
, (r>b)
3. Consider 3 straight, infinitely long, equally spaced wires (with zero radius), each carrying a current I in the
same direction.
(a)
Calculate the location of the two zeros in the magnetic field.
(b) Sketch the magnetic field line pattern
If the middle wire is rigidly displaced a very small distance x (x << d) upward are held fixed, describe
qualitatively the subsequent motion of the middle wire
Solution.
Assume the three wires are coplanar, then the points of zero magnetic field must also be located in the same
plane. Let the distance of such a point from the middle wire be x. Then the distance of this point from the other
two wires arc d±x. Applying Ampere’s circuital law we obtain for a point of zero magnetic field
$%&
=
&
+
$%&
Two solutions are possible, namely:
X='
(
)
Corresponding to two points located between the middle wire and each of the other 2 wires, both having
distance
(
)
from the middle wire.
(b) the magnetic field lines are as shown in figure 2.
Figure 2.
c. When the middle wire is displaced a small distance x in the same plane, the resultant force per unit length on
the wire is
f=
$%&
-
$!&
As x<<d, this force is approximately
f*
$
x
That is, the force is proportional but opposite to the displacement. Hence, the motion is simple harmonic about
the equilibrium position with a period T=
+
,
-
$
.
, where m is the mass per unit length of the middle wire.
This however is only one of the normal modes of oscillation of the middle wire. The other normal mode is
obtained when the wire is displaced a small distance x out from and normal to the plane as shown in figure 3.
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The resultant force on the wire is in the negative x direction, being
F=-2
/
$
%&
&
/
$
%&
*
$
0
This motion is also simple harmonic with the same period.
4. As in Figure 4, an infinitely long wire carries a current I = 1 A. It is bent so as to have a semi-circular detour
around the origin, with radius 1 cm. Calculate the magnetic field at the origin.
Figure 4.
Solution:
The straight parts of the wire do not contribute to the magnetic field at 0 since for them IdlXr = 0. We need only
to consider the contribution of the semi-circular part. The magnetic field at O produced by a current element Idl
is
dB=
1
$.2
3
With 1= 1 A, r = 10
2
m, the magnetic induction at O is B = 3.14 x 10
-5
T, pointing perpendicularly into the page
(Evaluate to get this answer!).
5. A semi-infinite solenoid of radius R and n turns per unit length carries a current I. Find an expression for the
radial component of the magnetic field B
r
(z
o
) near the axis at the end of the solenoid where r<< R and z = 0
6. Assume that the earth's magnetic field is caused by a. small current loop located at the center of the earth.
Given that the field near the pole is 0.8 gauss, that the radius of the earth is R = 6 x 10
6
m, and that 4
= 4
X
10
-7
H/m. use the Biot-Savart law to calculate the strength of the magnetic moment of the small current loop
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