The magnetic field on the axis of the magnet at a distance d1 = 14 cm, can be written as:
\(B_1\) = \(\frac{\mu_0 2M}{4\pi (d_1)^3}\)=H ...(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 d2, on the equatorial line of the magnet can be written as:
\(B_2\) = \(\frac{\mu_0 M}{4\pi (d_2)^3}\)=H ...(2)
Equating equations (1) and (2), we get:
\(\frac{2}{(d_1)^3}\)=\(\frac{1}{(d_2)^3}\)
\(\bigg(\frac{d_2}{d_1}\bigg)^3\)=\(\frac{1}{2}\)
∴ d2 = d1 × \(\bigg(\frac{1}{2}\bigg)^{\frac{1}{3}}\)
=14 × 0.794 = 11.1cm
The new null points will be located 11.1 cm on the normal bisector.
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).
Magnets are used in many devices like electric bells, telephones, radio, loudspeakers, motors, fans, screwdrivers, lifting heavy iron loads, super-fast trains, especially in foreign countries, refrigerators, etc.
Magnetite is the world’s first magnet. This is also called a natural magnet. Though magnets occur naturally, we can also impart magnetic properties to a substance. It would be an artificial magnet in that case.
Read More: Magnetism and Matter