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 (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 = H_N + H_S

= 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