Question

A bar magnet $30cm$ long is placed in the magnetic meridian with its north pole pointing south. The neutral point is observed at a distance of $30cm$ from its one end. Calculate the pole strength of the magnet. Given horizontal component of earth’s field is ${B_H} = 0.34G$
(A) $4.3Am$
(B) $5.2Am$
(C) $6.9Am$
(D) $8.6Am$

Answer 1

Hint: In order to solve this question, you need to know the terms used here and also know what is to be found out. The magnetic meridian is the line which passes through the north and south poles. A neutral point is a point where the net magnetic field due to magnet and earth is equal to zero. Pole strength is defined as the magnetic moment divided by the magnetic length. Use the appropriate formula in order to find the pole strength of the magnet.

Complete step by step answer:
When a magnet is placed in such a way that its north pole points towards the south, the neutral point is obtained at the axis of the magnet and as the point is said to be a neutral point, the magnitude of earth’s magnetic field is equal to the magnetic field of the magnet.
Therefore, we have, ${B_{magnet}} = {B_H}$

The magnetic field of the magnet at an axial point is given as
${B_{magnet}} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2Md}}{{{{\left( {{d^2} - {l^2}} \right)}^2}}}$ where $M$ is the magnetic moment.
Given: $d = 30cm = 03m$, $2l = 30cm \to l = 0.15m$ and ${B_H} = 0.34G = 0.34 \times {10^{ - 4}}T$.Let us substitute these values and find the magnetic moment.
\[
M = \left( {{B_{magnet}}} \right)\left( {\dfrac{{4\pi }}{{{\mu _0}}}} \right)\left( {\dfrac{{{{\left( {{d^2} - {l^2}} \right)}^2}}}{{2d}}} \right) \\
\Rightarrow M = \left( {{B_H}} \right)\left( {\dfrac{{4\pi }}{{{\mu _0}}}} \right)\left( {\dfrac{{{{\left( {{d^2} - {l^2}} \right)}^2}}}{{2d}}} \right) \\
\Rightarrow M = \left( {0.34 \times {{10}^{ - 4}}} \right)\left( {\dfrac{1}{{{{10}^{ - 7}}}}} \right)\left( {\dfrac{{{{\left( {{{0.3}^2} - {{0.15}^2}} \right)}^2}}}{{2 \times 0.3}}} \right) \\
\Rightarrow M = 2.582A{m^2} \\ \]
So, pole strength of the magnet will be,
$\therefore m = \dfrac{M}{{2l}} = \dfrac{{2.582}}{{0.3}} = 8.607Am$
Therefore, the magnetic pole strength of the given magnet is $8.6Am$.

Hence, option D is correct.

Note:You should always keep in mind the value of the quantity $\dfrac{{{\mu _0}}}{{4\pi }}$ which is $1 \times {10^{ - 7}}$. Also remember that the neutral point is defined as the point where the net magnetic field due to magnet and earth is zero. In case where the north pole of the magnet points towards the south pole of the earth, the magnetic field along the axis of the magnet will be given by ${B_{magnet}} = \dfrac{{{\mu _0}}}{{4\pi }}\dfrac{{2Md}}{{{{\left( {{d^2} - {l^2}} \right)}^2}}}$. Also remember to convert the units of the given quantities appropriately.