# 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. - Physics

Sum

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

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 15 Magnetic Properties of Matter
Q 4 | Page 286