Question

Two short bar magnets of length 1 cm each have magnetic moments 1.30A${{m}^{2}}$ and 1.00A${{m}^{2}}$ respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the south. They have a common magnetic equator and are separated by a distance of 20.0cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centers is close to
(Horizontal component if earth’s magnetic induction is $3.6\times {{10}^{-5}}Wb/{{m}^{2}}$ )
(A)$2.56\times {{10}^{-4}}Wb/{{m}^{2}}$
(B)$3.50\times {{10}^{-4}}Wb/{{m}^{2}}$
(C )5.80$\times {{10}^{-4}}Wb/{{m}^{2}}$
(D)$3.6\times {{10}^{-5}}Wb/{{m}^{2}}$

Answer 1

First calculate the magnetic field along the equator. Using the value of two magnetic moments calculate ${{B}_{1}} and {{B}_{2}}$ . Then find the resultant horizontal magnetic induction using the first and second value of the magnetic field. The horizontal component of earth’s magnetic field is the projection of earth’s magnetic field. The horizontal component of earth’s magnetic field of earth is measured using a compass.
Formula used:
$B=\dfrac{{{\mu }_{0}}m}{4\pi {{d}^{3}}}$

Complete step-by-step answer:

Given that,
$\begin{aligned} & {{m}_{1}}=1.20A{{m}^{2}} \\\ & {{m}_{2}}=1.00A{{m}^{2}} \\\ \end{aligned}$
B along equator,
$B=\dfrac{{{\mu }_{0}}m}{4\pi {{d}^{3}}}$
Here,
${{B}_{1}}=\dfrac{{{\mu }_{0}}}{4\pi }\times \dfrac{{{m}_{1}}}{{{d}^{3}}}$
Thus substituting the values in the given equation,
$\begin{aligned} & {{B}_{1}}=\dfrac{{{\mu }_{0}}}{4\pi }\times \dfrac{1.20}{{{10}^{-3}}}=1.20\times {{10}^{-4}}Wb/{{m}^{2}} \\\ & \\\ \end{aligned}$
Similarly,
${{B}_{2}}=\dfrac{{{\mu }_{0}}}{4\pi }\times \dfrac{{{m}_{2}}}{{{d}^{3}}}$
Then by substituting the values we get,
${{B}_{2}}=\dfrac{{{\mu }_{0}}}{4\pi }\times \dfrac{1}{{{10}^{-3}}}={{10}^{-4}}Wb/{{m}^{2}}$
$\theta ={{0}^{\circ }}$
Therefore B resultant is,
$B=\sqrt{B_{1}^{2}+B_{2}^{2}+2{{B}_{1}}{{B}_{2}}}$
Substituting the values we get,
$B=\left( \sqrt{{{1.2}^{2}}+{{1}^{2}}+2\times 1.2} \right)\times {{10}^{-4}}$
$=\left( \sqrt{1.44+1+2.4} \right)\times {{10}^{-4}}$
$=\sqrt{4.84}\times {{10}^{-4}}$
Thus,
$B=2.56\times {{10}^{-4}}Wb/{{m}^{2}}$

Hence option (A) is correct.

Additional information:
Magnetic induction also known as magnetic flux density describes a magnetic force on a test object like a small piece of iron in space. The horizontal component of earth’s magnetic field is the projection of earth’s magnetic field. The horizontal component of earth’s magnetic field of earth is measured using a compass. The needle of a compass is a small magnet that will always align with an external magnetic field. The opposite poles of the magnet attract and like poles repel.

Note:
The horizontal component of earth’s magnetic field is the projection of earth’s magnetic field. The horizontal component of earth’s magnetic field of earth is measured using a compass. The needle of a compass is a small magnet that will always align with an external magnetic field. The opposite poles of the magnet attract and like poles repel.