Question

A bar magnet of length 1 cm and cross-sectional area 1.0 cm² produces a magnetic field of 1.5 × 10⁻⁴ T at a point in end-on position at a distance 15 cm away from the centre. (a) Find the magnetic moment M of the magnet. (b) Find the magnetisation I of the magnet. (c) Find the magnetic field B at the centre of the magnet.

Answer 1

Given:
Distance of the observation point from the centre of the bar magnet, d = 15 cm = 0.15 m
Length of the bar magnet, l = 1 cm = 0.01 m
Area of cross-section of the bar magnet, A = 1.0 cm² = 1 × 10⁻⁴ m²
Magnetic field strength of the bar magnet, B = 1.5 × 10⁻⁴ T
As the observation point lies at the end-on position, magnetic field $\left(B\right)$ is given by,
$\overrightarrow{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{2Md}{(d^2-l^2{)}^2}$
On substituting the respective values, we get:
$$\begin{gathered} 1.5\times {10}^{-4}=\frac{{10}^{-7}\times 2\times M\times 0.15}{(0.0225-0.0001{)}^2} \\ \Rightarrow 1.5\times {10}^{-4}=\frac{3\times {10}^{-8}\times M}{5.01\times {10}^{-4}} \\ \Rightarrow M=\frac{1.5\times {10}^{-4}\times 5.01\times {10}^{-4}}{3\times {10}^{-8}} \\ =2.5\mathrm{A} \end{gathered}$$
(b) Intensity of magnetisation (I) is given by,
I = $\frac{M}{V}$
$$\begin{gathered} =\frac{2.5}{{10}^{-4}\times {10}^{-2}} \\ =2.5\times {10}^6\mathrm{A}/\mathrm{m} \end{gathered}$$
(c)
$\mathrm{H}=\frac{\mathrm{M}}{4\mathrm{\pi }l{d}^2}$
$$\begin{gathered} =\frac{2.5}{4\times 3.14\times 0.01\times (0.15{)}^2} \\ =\frac{2.5}{4\times 3.14\times 1\times {10}^{-2}\times 2.25\times {10}^{-2}} \end{gathered}$$
Net H = H_N + H_S
= 884.6 = 8.846 × 10²
= 314 T
$\overrightarrow{\mathrm{B}}$ = µ₀ (H + 1)
= π × 10⁻⁷ (2.5 × 10⁶ + 2 × 884.6)
= 3.14 T