Question
If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points be located?
✓
Answer
The rnaqnettc field on the axis of the rnagnet at a distance d1 = 14 cm, can be written as:
$\text{B}_1=\frac{\mu_02\text{M}}{4\pi(\text{d}_1)^3}=\text{H}\ \dots(1)$
Where,
M = Magnetic moment
$\mu_0$ = Permeability of free space
H = Horizontal component of the magnetic field at d1
If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.
Hence, the magnetic field at a distance dz, on the equatorial line of the magnet can be Written as:
$\text{B}_2=\frac{\mu_0\text{M}}{4\pi(\text{d}_2)^3}=\text{H}\ \dots(2)$
Equating equations (1) and (2), we get:
$\frac{2}{(\text{d}_1)^3}=\frac{1}{(\text{d}_2)^3}$
$\Big(\frac{\text{d}_2}{\text{d}_1}\Big)^3=\frac{1}{2}$
$\therefore\ \text{d}_2=\text{d}_1\times\Big(\frac{1}{2}\Big)^{\frac{1}{3}}$
= 14 × 0794 = 11.1 cm
The new null points will be located 11.1 cm on the normal bisector.
$\text{B}_1=\frac{\mu_02\text{M}}{4\pi(\text{d}_1)^3}=\text{H}\ \dots(1)$
Where,
M = Magnetic moment
$\mu_0$ = Permeability of free space
H = Horizontal component of the magnetic field at d1
If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.
Hence, the magnetic field at a distance dz, on the equatorial line of the magnet can be Written as:
$\text{B}_2=\frac{\mu_0\text{M}}{4\pi(\text{d}_2)^3}=\text{H}\ \dots(2)$
Equating equations (1) and (2), we get:
$\frac{2}{(\text{d}_1)^3}=\frac{1}{(\text{d}_2)^3}$
$\Big(\frac{\text{d}_2}{\text{d}_1}\Big)^3=\frac{1}{2}$
$\therefore\ \text{d}_2=\text{d}_1\times\Big(\frac{1}{2}\Big)^{\frac{1}{3}}$
= 14 × 0794 = 11.1 cm
The new null points will be located 11.1 cm on the normal bisector.
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