Answer
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Hint- Here, we will proceed by expressing the magnetic pole strength mathematically in terms of quantities whose units in the M.K.S. system (or SI units) are known well before. Finally, we will be putting these units in the formula obtained.
Formulas Used- ${\text{B}} = \dfrac{{\text{F}}}{{\text{m}}}$, $F = qvB$ and ${\text{I}} = \dfrac{{\text{q}}}{{\text{t}}}$.
Complete step-by-step solution -
Here, it is important to note that the M.K.S. system is simply the SI system. This means that the unit of magnetic pole strength in the M.K.S. system is the same as that in the SI system.
As we know that the magnetic field can be mathematically written as the ratio of the magnetic force and the magnetic pole strength
i.e., ${\text{B}} = \dfrac{{\text{F}}}{{\text{m}}}{\text{ }} \to \left( 1 \right)$ where B denotes the magnitude of magnetic field, F denotes the magnetic force and m denotes the magnetic pole strength
By shifting m from the RHS to the LHS and B from the LHS to the RHS in equation (1), we get
${\text{m}} = \dfrac{{\text{F}}}{{\text{B}}}{\text{ }} \to \left( 2 \right)$
Also, we know that the magnetic force can be written as
F = qvB where F denotes the magnetic force induced in a conductor, q denotes the charge, v denotes the velocity with which conductor is moving and B denotes as the magnitude of the magnetic field
By substituting F = qvB in equation (2), we get
\[
\Rightarrow {\text{m}} = \dfrac{{{\text{qvB}}}}{{\text{B}}} \\
\Rightarrow {\text{m}} = {\text{qv }} \to \left( 3 \right) \\
\]
Also, we know that current I is defined as charge flowing q per unit time t
i.e., $
{\text{I}} = \dfrac{{\text{q}}}{{\text{t}}} \\
\Rightarrow {\text{q}} = {\text{It}} \\
$
By substituting q = It in equation (3), we get
\[ \Rightarrow {\text{m}} = {\text{Itv }} \to {\text{(4)}}\]
As, Unit of the current (I) = Amperes (A), Unit of time t = seconds (s) and Unit of velocity v = metre per second $\dfrac{{\text{m}}}{{\text{s}}}$
Using equation (4), we can write
Unit of magnetic pole strength m = (Unit of current I)( Unit of time t)( Unit of velocity v)
$ \Rightarrow $ Unit of magnetic pole strength m = ${\text{A}} \times {\text{s}} \times \dfrac{{\text{m}}}{{\text{s}}} = {\text{Am}}$
Therefore, the unit of magnetic pole strength in the M.K.S. system (or SI units) is Ampere-metre (Am).
Note- The strength of a magnetic pole to attract other magnetic materials towards itself is called magnetic pole strength. The magnetic pole strength is a scalar quantity (i.e., only magnitude is there). Usually, the pole strengths of north and south poles of a magnet are represented by +m and –m respectively.
Formulas Used- ${\text{B}} = \dfrac{{\text{F}}}{{\text{m}}}$, $F = qvB$ and ${\text{I}} = \dfrac{{\text{q}}}{{\text{t}}}$.
Complete step-by-step solution -
Here, it is important to note that the M.K.S. system is simply the SI system. This means that the unit of magnetic pole strength in the M.K.S. system is the same as that in the SI system.
As we know that the magnetic field can be mathematically written as the ratio of the magnetic force and the magnetic pole strength
i.e., ${\text{B}} = \dfrac{{\text{F}}}{{\text{m}}}{\text{ }} \to \left( 1 \right)$ where B denotes the magnitude of magnetic field, F denotes the magnetic force and m denotes the magnetic pole strength
By shifting m from the RHS to the LHS and B from the LHS to the RHS in equation (1), we get
${\text{m}} = \dfrac{{\text{F}}}{{\text{B}}}{\text{ }} \to \left( 2 \right)$
Also, we know that the magnetic force can be written as
F = qvB where F denotes the magnetic force induced in a conductor, q denotes the charge, v denotes the velocity with which conductor is moving and B denotes as the magnitude of the magnetic field
By substituting F = qvB in equation (2), we get
\[
\Rightarrow {\text{m}} = \dfrac{{{\text{qvB}}}}{{\text{B}}} \\
\Rightarrow {\text{m}} = {\text{qv }} \to \left( 3 \right) \\
\]
Also, we know that current I is defined as charge flowing q per unit time t
i.e., $
{\text{I}} = \dfrac{{\text{q}}}{{\text{t}}} \\
\Rightarrow {\text{q}} = {\text{It}} \\
$
By substituting q = It in equation (3), we get
\[ \Rightarrow {\text{m}} = {\text{Itv }} \to {\text{(4)}}\]
As, Unit of the current (I) = Amperes (A), Unit of time t = seconds (s) and Unit of velocity v = metre per second $\dfrac{{\text{m}}}{{\text{s}}}$
Using equation (4), we can write
Unit of magnetic pole strength m = (Unit of current I)( Unit of time t)( Unit of velocity v)
$ \Rightarrow $ Unit of magnetic pole strength m = ${\text{A}} \times {\text{s}} \times \dfrac{{\text{m}}}{{\text{s}}} = {\text{Am}}$
Therefore, the unit of magnetic pole strength in the M.K.S. system (or SI units) is Ampere-metre (Am).
Note- The strength of a magnetic pole to attract other magnetic materials towards itself is called magnetic pole strength. The magnetic pole strength is a scalar quantity (i.e., only magnitude is there). Usually, the pole strengths of north and south poles of a magnet are represented by +m and –m respectively.
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