Question (2ha0001)
Deection of a compass needle by a dipole magnet
A dipole magnet and compass are arranged as shown below, with the dipole aligned East-West with the
compass. The dipole moment of the magnet is 1.5 A · m2 and its center is 20 cm from the center of a
compass. The magnetic eld of Earth is shown.
(a) Sketch and calculate the deection of the compass needle from North.
(b) If you replace the magnetic dipole with a thin coil of radius 2 cm and 20 turns at the same location,
what must be the current in the coil to give the same deection of the compass needle as the dipole
magnet?
Figure 1: Dipole magnet and compass.
Solution
The magnetic eld at the location of the compass points toward the S pole of the dipole, and Earth's
magnetic eld points North (that is, approximately geographic North) as shown below.
(a)
Figure 2: Magnetic elds at the location of the compass.
Thus, the compass will deect to the left, or toward the West side of the compass, as shown below.
The horizontal component of the magnetic eld of Earth at the latitude of High Point, NC, is approximately
22,000 nT, or 2.2 × 10−5 T. (You can calculate the horizontal component of the magnetic eld of Earth at
your latitude at <http://www.ngdc.noaa.gov/geomagmodels/IGRFWMM.jsp> . The magnetic eld at the
location of the compass due to the dipole has a magnitude
Figure 3: Deection of the compass needle.
Bdipole,axis
≈
µ0 2µ
4π r3
≈ (1 × 10−7 T m2 /(C m/s))
2(1.5 A · m2 )
(0.2 m)3
≈ 3.75 × 10−5 T
The components of the net magnetic eld are the magnetic eld of the dipole and the magnetic eld of
Earth. They can be used to nd the deection angle, shown below.
Figure 4: The net magnetic eld and the deection angle of the compass needle.
From the right triangle, the deection angle is computed by
θ
Bdipole
Bearth
3.75 × 10−5 T
=
2.2 × 10−5 T
3.75 × 10−5 T
−1
= tan
2.2 × 10−5 T
◦
= 59.6
θ
≈ 60◦
tan θ
θ
=
If you replace the dipole magnet with a coil, then the current in the coil will have to ow around the
coil in such a way that it creates a magnetic eld along the axis of the coil that is to the left. Using the
right hand rule, with your thumb pointing left, your ngers go into the screen at the top of the coil and out
of the screen at the bottom of the coil. A side view of the coil is shown below.
The dipole moment of a coil of N turns is
(b)
Figure 5: Orientation of a current-carrying coil with the same dipole moment as the previously used magnet.
µcoil = N IA
which must be the same as the previously used magnet of dipole moment µ = 1.5 A m2 . Thus,
I
=
=
=
=
µ
NA
µ
N πR2
1.5 A m2
20π(0.022 )
60 A
Wow, 60 A is a lot of current going through that wire coil! One way to reduce the current is to increase the
number of turns in the coil. Also, one can make a coil with a larger radius.