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.
Solution
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 (B) is given by,
\[\overrightarrow{B} = \frac{\mu_0}{4\pi} \times \frac{2Md}{( d^2 - l^2 )^2}\]
On substituting the respective values, we get:-
\[1 . 5 \times {10}^{- 4} = \frac{{10}^{- 7} \times 2 \times M \times 0 . 15}{(0 . 0225 - 0 . 0001 )^2}\]
\[ \Rightarrow 1 . 5 \times {10}^{- 4} = \frac{3 \times {10}^{- 8} \times M}{5 . 01 \times {10}^{- 4}}\]
\[ \Rightarrow M = \frac{1 . 5 \times {10}^{- 4} \times 5 . 01 \times {10}^{- 4}}{3 \times {10}^{- 8}}\]
\[= 2 . 5 A\]
(b) Intensity of magnetisation (I) is given by,
`I = M/V`
\[= \frac{2 . 5}{{10}^{- 4} \times {10}^{- 2}}\]
\[ = 2 . 5 \times {10}^6 \text{ A/m}\]
(c) \[H = \frac{M}{4\pi ld^2}\]
\[= \frac{2 . 5}{4 \times 3 . 14 \times 0 . 01 \times (0 . 15 )^2}\]
\[ = \frac{2 . 5}{4 \times 3 . 14 \times 1 \times {10}^{- 2} \times 2 . 25 \times {10}^{- 2}}\]
Net H = HN + HS
= 884.6 = 8.846 × 102
= 314 T
\[\overrightarrow{B}=mu_0\left(H+1\right)\]
= π × 10−7 (2.5 × 106 + 2 × 884.6)
= 3.14 T
