A magnet of magnetic moment m is first cut perpendicular to its length and then perpendicular to its breadth into n equal parts, then calculate magnetic moment of each part.
Answer
Magnetic moment of each part $=\frac{m}{n} \times \frac{ L }{n}=\frac{m L}{n^2}$
Magnetic moment ( m ) generated in unit volume of a substance is called intensity of magnetisation. $\overrightarrow{I}=\frac{\overrightarrow{m}}{V}$ It is a vector quantity and its unit is $A / m$.
Such points where the resultant magnetic field due to the magnetic field of the magnet and the horizontal component of the earth's magnetic field is zero are known as neutral points.
Since there is no existence of isolated pole, that is, poles always exist in pairs. Magnetic lines of force emerge from the north pole and move inside the south pole, hence they make closed curves.
What is the ratio of the intensities of the points situtated at equal distances from the midpoint of a short bar magnet on its axial and non-axial points?
Write an expression for intensity of magnetisation field due to a small bar magnet of magnetic moment $\overrightarrow{ M }$ at a distance $r$ on its non-axial side.
Answer
$\overrightarrow{ B }=\frac{\mu_0}{4 \pi} \times\left(\frac{-\overrightarrow{ m }}{ r ^3}\right)$
Write an expression for intensity of magnetisation field at a point located at a distance $r$ on its axis due to a small bar magnet having magnetic moment $\overrightarrow{ m }$.
Product of effective length and pole strength of a magnet is known as magnetic moment. It is a vector quantity whose direction is from south pole to the north pole. Its unit is ampere-metre ${ }^2$.
How should a magnetic dipole be placed in a uniform magnetic field so that its potential energy is maximum?
Answer
Its magnetic moment should be in opposite direction of $\overrightarrow{ m } \cdot \overrightarrow{ B }$ $\because$ In this condition, $\theta=180^{\circ}$ and $U =-m B \cos \theta$ $U =-m B \cos 180^{\circ}=-m B \times(-1)$ $U =m B$
Name the physical quantity whose unit is newton metre tesla ${ }^{-1}$ and whether is it scalar or vector?
Answer
$\tau_{\max }=m B \Rightarrow m=\frac{\tau_{\max }}{ B }$ On the basis of this formula, S.I. unit of M is newton metre tesla ${ }^{-1}$. Hence, newton metre tesla ${ }^{-1}$ is S.I. unit of magnetic moment. It is a vector quantity.