Question

A bar magnet $20\,{\text{cm}}$ in length is placed with its south pole towards geographic north. The neutral points are situated at a distance of $40\,{\text{cm}}$ from the centre of the magnet. If horizontal component of earth’s field $3.2 \times {10^{ - 5\,}}\,{\text{T}}$, then pole strength of magnet is:
A. $5\,{\text{Am}}$
B. $10\,{\text{Am}}$
C. $45\,{\text{Am}}$
D. $20\,{\text{Am}}$

Answer 1

Use the formula for the magnetic field of a bar magnet at an axial point and the formula for the magnetic moment of the magnet in terms of pole strength of the magnet.

Formula used:
The formula for the magnetic field $B$ of a bar magnet at an axial point is
$B = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2Md}}{{{{\left( {{d^2} - {l^2}} \right)}^2}}}$ …… (1)
Here, ${\mu _0}$ is absolute permeability of free space, $M$ is the magnetic moment of the magnet, $d$ is the distance between the centre of the magnet and the neutral point and $l$ is the half length of the magnet.
The magnetic moment $M$ of the magnet is given by
$M = m\left( {2l} \right)$ …… (2)
Here, $m$ is the pole strength of the magnet and $2l$ is the length of the magnet.

Complete step by step answer:
The length of the bar magenta is $20\,{\text{cm}}$.
$2l = 20\,{\text{cm}}$
$\Rightarrow l = 10\,{\text{cm}}$
The neutral points are situated at a distance of $40\,{\text{cm}}$ from the centre of the magnet.
$d = 40\,{\text{cm}}$
The horizontal component of earth’s field is $3.2 \times {10^{ - 5\,}}\,{\text{T}}$.
${B_H} = 3.2 \times {10^{ - 5\,}}\,{\text{T}}$
Rewrite equation (1) for the horizontal magnetic field of the Earth for given the bar magnet.
${B_H} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2Md}}{{\left( {{d^2} - {l^2}} \right)}}$
Rearrange the above equation for $M$.
$M = \dfrac{{4\pi }}{{2{\mu _0}d}}{\left( {{d^2} - {l^2}} \right)^2}{B_H}$
Substitute ${10^{ - 7}}\,{\text{H/m}}$ for $\dfrac{{{\mu _0}}}{{4\pi }}$, $40\,{\text{cm}}$ for $d$, $10\,{\text{cm}}$ for $l$ and $3.2 \times {10^{ - 5\,}}\,{\text{T}}$ for ${B_H}$ in the above equation.
$M = \dfrac{1}{{2\left( {{{10}^{ - 7}}\,{\text{H/m}}} \right)\left( {40\,{\text{cm}}} \right)}}{\left[ {{{\left( {40\,{\text{cm}}} \right)}^2} - {{\left( {10\,{\text{cm}}} \right)}^2}} \right]^2}\left( {3.2 \times {{10}^{ - 5\,}}\,{\text{T}}} \right)$
$\Rightarrow M = \dfrac{1}{{2\left( {{{10}^{ - 7}}\,{\text{H/m}}} \right)\left[ {\left( {40\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)} \right]}}{\left[ {{{\left[ {\left( {40\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)} \right]}^2} - {{\left[ {\left( {10\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)} \right]}^2}} \right]^2}\left( {3.2 \times {{10}^{ - 5\,}}\,{\text{T}}} \right)$
$\Rightarrow M = 9\,{\text{A}} \cdot {{\text{m}}^2}$
Hence, the magnetic moment of the magnet is $9\,{\text{A}} \cdot {{\text{m}}^2}$.
Determine the pole strength of the magnet.
Rearrange equation (2) for the pole strength of the magnet.
$m = \dfrac{M}{{2l}}$
Substitute $9\,{\text{A}} \cdot {{\text{m}}^2}$ for $M$ and $20\,{\text{cm}}$ for $2l$ in the above equation.
$m = \dfrac{{9\,{\text{A}} \cdot {{\text{m}}^2}}}{{20\,{\text{cm}}}}$
$\Rightarrow m = \dfrac{{9\,{\text{A}} \cdot {{\text{m}}^2}}}{{\left( {20\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}\,}}} \right)}}$
$\Rightarrow m = 45\,{\text{A}} \cdot {\text{m}}$
Therefore, the pole strength of the pole is $45\,{\text{A}} \cdot {\text{m}}$.

So, the correct answer is “Option C”.

Note:
Convert the units of the length of the magnet and the distance of the centre of magnet from the neutral point in the SI system of units.
The magnetic moment of the magnet in terms of pole strength of the magnet.