Obtain an expression for the magnetic field due to a short bar magnet, at a point on its axis at a distance x from its centre considering it as an equivalent of a solenoid.
The strength of magnetic field at
any point around a bar magnet can be calculated. However, the measurements at
two points are important: at a point on the axis and at a point on the
equatorial line of the bar magnet. These are called end-on position and broadside-on
position respectively.
Consider a bar magnet of length 2l
and pole strength m. Suppose a point P on the axis of the magnet at a distance
d from its center. (d –l) is the distance of P form the N-pole of the magnet.
The magnetic field intensity at P due to the north-pole of the magnet is
B1=μ04πmr2=μ04πμ0m(d–l)2
which is directly away from N-pole.
Since the south of the magnet is at a distance r = d + l from P, so magnetic
field intensity at P due to S-pole is
B2=μ04πmr2=μ04πμ0m(d+l)2
which is direct towards, the S-pole
of the magnet. The magnetic field intensity B at P is the resultant of these
two fields,
B=B1+(−B2)=B1–B2=μ0m4π[1(d–l)2−1(d+l)2]=μ0m4π[4ld(d2–l2)2]=μ02md4π(d2–l2)2
where M=m×2, the magnetic
moment of the bar magnet. So, the magnetic field at a point on the axis of a
bar magnet is
B=μ0md2π(d2–l2)2
If the length of the magnet is very
small, d>>I and the magnetic field intensity is
B=μ0m2πd3
Suppose a point P is on the
equatorial line of the bar magnet. The equatorial line of the magnet is the
line perpendicular to the axis of the magnet which bisects the magnet. Let d be
the distance of the point P from the centre of the magnet and P due to the
North Pole is
B1=μ04πmr2=μ04πμ0m(d–l)2
B2=μ04πmr2=μ04πμ0m(d+l)2
directed towards S-pole. These
fields have different directions, but the same magnitude as shown in the
figure. Let ∠PSO=θ∠PSO=θ and by symmetry, ∠PNO=θ∠PNO=θ. The angle between B1 and B2 is then 2θ2θ. The resultant
magnetic field, B at P is given by
The direction of B is parallel to
the axis of the magnet, from north to south pole. If the magnet is very short, d>>ld>>l, and the magnetic
field at P is
B=μ04πmd3
The strength of magnetic field at any point around a bar magnet can be calculated. However, the measurements at two points are important: at a point on the axis and at a point on the equatorial line of the bar magnet. These are called end-on position and broadside-on position respectively.
Consider a bar magnet of length 2l and pole strength m. Suppose a point P on the axis of the magnet at a distance d from its center. (d –l) is the distance of P form the N-pole of the magnet. The magnetic field intensity at P due to the north-pole of the magnet is
B1=μ04πmr2=μ04πμ0m(d–l)2
which is directly away from N-pole. Since the south of the magnet is at a distance r = d + l from P, so magnetic field intensity at P due to S-pole is
B2=μ04πmr2=μ04πμ0m(d+l)2
which is direct towards, the S-pole of the magnet. The magnetic field intensity B at P is the resultant of these two fields,
B=B1+(−B2)=B1–B2=μ0m4π[1(d–l)2−1(d+l)2]=μ0m4π[4ld(d2–l2)2]=μ02md4π(d2–l2)2
where M=m×2, the magnetic moment of the bar magnet. So, the magnetic field at a point on the axis of a bar magnet is
B=μ0md2π(d2–l2)2
If the length of the magnet is very small, d>>I and the magnetic field intensity is
B=μ0m2πd3
Suppose a point P is on the equatorial line of the bar magnet. The equatorial line of the magnet is the line perpendicular to the axis of the magnet which bisects the magnet. Let d be the distance of the point P from the centre of the magnet and P due to the North Pole is
B1=μ04πmr2=μ04πμ0m(d–l)2
B2=μ04πmr2=μ04πμ0m(d+l)2
directed towards S-pole. These fields have different directions, but the same magnitude as shown in the figure. Let ∠PSO=θ∠PSO=θ and by symmetry, ∠PNO=θ∠PNO=θ. The angle between B1 and B2 is then 2θ2θ. The resultant magnetic field, B at P is given by
The direction of B is parallel to the axis of the magnet, from north to south pole. If the magnet is very short, d>>ld>>l, and the magnetic field at P is
B=μ04πmd3
