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

Answer:

:

Explanation:

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Answer 2

(a) The magnetic moment M of the magnet is $2.5 \mathrm{A}$

(b) The magnetisation I of the magnet is  $2.5 \times 10^{6} \frac{A}{m}$

(c) The magnetic field B at the centre of the magnet is $3.14 T$

Explanation:

(a)The magnetic moment M of the magnet:

Step 1:

Observation point distance from the bar magnet core, d = 15 cm = 0.15 m

Bar magnet longitude, l = 1 cm = 0.01 m

Cross-section area of magnet bar, A = 1.0 $cm^{2}$ = $1 \times 10^{-4} \mathrm{m}^{-2}$

Bar magnet's strength in magnetic field, B = 1.5 × 10−4 T

Step 2:

Since the point of observation is in the end-on position, the magnetic field is

$\vec{B}=\frac{\mu_{0}}{4 \pi} \times \frac{2 M d}{\left(d^{2}-l^{2}\right)^{2}}$

Step 3:

When the respective values are replaced we get:

$1.5 \times 10^{-4}=10^{-7} \times 2 \times M \times \frac{0.15}{\left(0.15^{2}-0.01^{2}\right)^{2}}$

$1.5 \times 10^{-4}=10^{-7} \times 2 \times M \times \frac{0.15}{(0.0225-0.0001)^{2}}$

$1.5 \times 10^{-4}=3 \times 10^{-8} \times \frac{M}{\left(5.01 \times 10^{-4}\right)}$  

           $M=1.5 \times 10^{-4} \times 5.01 \times \frac{10^{-4}}{\left(3 \times 10^{-8}\right)}$

                = 2.5 A

(b)The magnetisation I of the magnet :

The magnetisation intensity I is given by,

$\mathrm{I}=\frac{M}{V}$

 $=\frac{2.5}{\left(10^{-4} \times 10^{-2}\right)}$

 $=2.5 \times 10^{6} \frac{A}{m}$

(c)  The magnetic field B at the centre of the magnet:

$H=\frac{M}{4 \pi l d^{2}}$

   $=\frac{2.5}{4 \times 3.14 \times 0.01 \times(0.15)^{2}}$

  $=\frac{2.5}{4 \times 3.14 \times 1 \times 10^{-2} \times 2.25 \times 10^{-2}}$

$\text { Net } H=H_{N}+H_{S}$

= 884.6 = 8.846 × 〖10〗^2

= 314 T

$\vec{B}=m \mu_{0}(H+1)$

$\begin{aligned}&=\pi \times 10^{-7}\left(2.5 \times 10^{6}+2 \times 884.6\right)\\&=3.14 T\end{aligned}$