Obtain an expression for Magnetic field along the equatorial point.
Answers
Answer:
Consider a point P on equatorial (or broad side on position) of short bar magnet of length 2I, having north pole (N) and south pole (S) of strength +q
m
and −q
m
respectively. The distance of point P from the mid-point (O) of magnet is r. Let B
1
and B
2
be the magnetic field intensities due to north and south poles respectively. NP=SP=
r
2
+l
2
.
vecB
1
=
4π
μ
0
r
2
+l
2
q
m
along N to P
vecB
2
=
4π
μ
0
r
2
+l
2
q
m
along P to S
Clearly, magnitude of and are equal
i.e., ∣
B
1
∣=∣
B
2
∣ or B
1
=B
2
To find the resultant of
B
1
and
B
2
we resolve them along and perpendicular to magnetic axis SN components
B
1
of along and perpendicular to magnetic axis are B
1
cosθ and B
2
sinθ respectively.
Components of
B
2
along and perpendicular to magnetic axis are 2cosθ and B2sinθ respectively. Clearly components of and perpendicular to axis SN. B1sinθ and B2sinθ ae equal in magnitude and opposite in direction and hence, cancel each other, while the components of
B1
and
B2
along the axis are in the same direction and hence, add up to give to resultant magnetic filed parallel to the direction
NS
.
∴ Resultant magnetic field intensity at P.
B=B
1
cosθ+B
2
cosθ
But B
1
=B
2
=
4π
μ
0
r
2
+l
2
q
m
ancosθ=
PN
ON
=
r
2
+l
2
1
=
(r
2
+l
2
)
1/2
1
∴ B=2B
1
cosθ=2×
4π
μ
0
(r
2
+l
2
)
q
m
×
(r
2
+l
2
)
1/2
l
=
4π
μ
0
(r
2
+l
2
)
3/2
2q
m
l
But q
m
2I=m, magnetic moment of magnet
∴ B=
4π
μ
0
(r
2
+l
2
)
3/2
m
If the magnet is very short and point P is far away, we have I<<r; so 12 may be neglect as compared to r
2
and so equation (3) takes the form
B
4π
μ
0
r
3
m
This is the expression for magnetic field intensity at the equatorial position of the magnet.
It is clear from equation (2) and (4) that the magnetic field strength due to a short magnetic dipole is inversely proportional to the cube of its distance from the center of the dipole and the magnetic field intensity at axial position is twice that at the equatorial position for the same distance.