
A compass needle placed at a distance r from a short magnet in a position shows deflection of ${{60}^{o}}$. If the distance is increased to $r{{\left( 3 \right)}^{\dfrac{1}{3}}}$, then deflection of compass needle, in degrees, is:
A. 30
B. $60\times {{3}^{^{\dfrac{1}{3}}}}$
C. $60\times {{3}^{^{\dfrac{2}{3}}}}$
D. $60\times 3$
Answer
585.6k+ views
Hint: The magnetic force of a magnetic field will follow inverse cube law. So, the magnetic field will be inversely proportional to the cube of distance from the centre of the bar magnet for a short bar magnet. We will take the magnetic force due to both the magnetic field of the earth and the bar magnet in consideration here when calculating the deflection of the needle of the compass.
Formula used:
$F\propto \dfrac{1}{{{r}^{3}}}$
Complete answer:
As a bar magnet is a magnetic dipole and as we have seen for electric dipoles, the force at a point in space is inversely proportional to the distance of the point from the centre of the dipole for a short dipole, i.e. where the distance between the two poles is very small when compared to the distance of the point from the centre. This is the case here and taking an analogy with the electric dipole we will take the force to follow the inverse cube law.
As we can see from the figure, ${{M}_{e}}$ is the force due to earth’s magnetic field and ${{M}_{b}}$ is the force due to the magnetic field of the bar magnet. We can also write that
$\tan 60=\dfrac{{{M}_{b}}}{{{M}_{e}}}=\sqrt{3}\Rightarrow {{M}_{b}}=\sqrt{3}\times {{M}_{e}}$
When the distance is increased to $r{{\left( 3 \right)}^{\dfrac{1}{3}}}$, let the force be ${{M}_{b}}^{\prime }$
\[\dfrac{{{M}_{b}}^{\prime }}{{{M}_{b}}}=\dfrac{{{r}^{3}}}{{{\left( r{{\left( 3 \right)}^{\dfrac{1}{3}}} \right)}^{3}}}=\dfrac{{{r}^{3}}}{3{{r}^{3}}}=\dfrac{1}{3}\Rightarrow {{M}_{b}}^{\prime }=\dfrac{{{M}_{b}}}{3}\]
And we get
\[\begin{align}
& 3{{M}_{b}}^{\prime }=\sqrt{3}\times {{M}_{e}} \\
& \Rightarrow \dfrac{{{M}_{b}}^{\prime }}{{{M}_{e}}}=\dfrac{1}{\sqrt{3}} \\
\end{align}\]
The angle of deflection will now be equal to \[{{\tan }^{-1}}\dfrac{1}{\sqrt{3}}={{30}^{o}}\]. Hence, the correct option is A, i.e. 30.
So, the correct answer is “Option A”.
Note:
Students can also derive the formula for magnetic dipole force, but it is easier to use the results from the formula for electric dipole as the proportionality is the same, although the proportionality constant changes. Also, it must be noted that the result derived is for a short dipole or a short bar magnet and should be used in only such conditions.
Formula used:
$F\propto \dfrac{1}{{{r}^{3}}}$
Complete answer:
As a bar magnet is a magnetic dipole and as we have seen for electric dipoles, the force at a point in space is inversely proportional to the distance of the point from the centre of the dipole for a short dipole, i.e. where the distance between the two poles is very small when compared to the distance of the point from the centre. This is the case here and taking an analogy with the electric dipole we will take the force to follow the inverse cube law.
As we can see from the figure, ${{M}_{e}}$ is the force due to earth’s magnetic field and ${{M}_{b}}$ is the force due to the magnetic field of the bar magnet. We can also write that
$\tan 60=\dfrac{{{M}_{b}}}{{{M}_{e}}}=\sqrt{3}\Rightarrow {{M}_{b}}=\sqrt{3}\times {{M}_{e}}$
When the distance is increased to $r{{\left( 3 \right)}^{\dfrac{1}{3}}}$, let the force be ${{M}_{b}}^{\prime }$
\[\dfrac{{{M}_{b}}^{\prime }}{{{M}_{b}}}=\dfrac{{{r}^{3}}}{{{\left( r{{\left( 3 \right)}^{\dfrac{1}{3}}} \right)}^{3}}}=\dfrac{{{r}^{3}}}{3{{r}^{3}}}=\dfrac{1}{3}\Rightarrow {{M}_{b}}^{\prime }=\dfrac{{{M}_{b}}}{3}\]
And we get
\[\begin{align}
& 3{{M}_{b}}^{\prime }=\sqrt{3}\times {{M}_{e}} \\
& \Rightarrow \dfrac{{{M}_{b}}^{\prime }}{{{M}_{e}}}=\dfrac{1}{\sqrt{3}} \\
\end{align}\]
The angle of deflection will now be equal to \[{{\tan }^{-1}}\dfrac{1}{\sqrt{3}}={{30}^{o}}\]. Hence, the correct option is A, i.e. 30.
So, the correct answer is “Option A”.
Note:
Students can also derive the formula for magnetic dipole force, but it is easier to use the results from the formula for electric dipole as the proportionality is the same, although the proportionality constant changes. Also, it must be noted that the result derived is for a short dipole or a short bar magnet and should be used in only such conditions.
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