Match List-I with List-II

$\mathrm{List-I}\left(\text{Physical Quantity}\right)$	$\mathrm{List-II}\left(\text{Units}\right)$
$\left(A\right)\text{Magnetic field}$	$\left(I\right)J{T}^{-}$
$\left(B\right)\text{Magnetic moment}$	$\left(\mathrm{II}\right)Tm{A}^{-}$
$\left(C\right)\text{Pole strength}$	$\left(\mathrm{III}\right)J{T}^{-}{m}^{-}$
$\left(D\right)\text{Permeability of free space}$	$\left(\mathrm{IV}\right)\mathrm{Wb}{m}^{-}$

Choose the correct answer from the options given below:

The correct option is (A)- (IV), (B)- (I), (C)- (III), (D)- (II).

Let us analyze each physical quantity in List-I and determine its corresponding SI unit in List-II.

(A) Magnetic field:
The magnetic field is defined as the magnetic flux per unit area.
The SI unit of magnetic flux is the Weber ($\mathrm{Wb}$), and the unit of area is square meters ($m^2$).
Therefore, the unit of magnetic field is:

$\text{Unit of Magnetic Field}=\frac{\mathrm{Wb}}{m^2}=\mathrm{Wb}{m}^{\&\#x2212;2}$

This matches with (IV) in List-II. Thus, (A) matches with (IV).

(B) Magnetic moment:
The potential energy ($U$) of a magnetic dipole in a magnetic field ($B$) is given by the relation:

$U=\&\#x2212;M\&\#x22C5;B\&\#x2061;\cos \left(\mathrm{\&\#x03B8;}\right)$

where $M$ is the magnetic moment. Rearranging for $M$, we find:

$M=\frac{U}{B\&\#x2061;\cos \left(\mathrm{\&\#x03B8;}\right)}$

Since energy is measured in Joules ($J$) and magnetic field in Tesla ($T$), the unit of magnetic moment is:

$\text{Unit of Magnetic Moment}=\frac{J}{T}=J{T}^{\&\#x2212;1}$

This matches with (I) in List-II. Thus, (B) matches with (I).

(C) Pole strength:
The magnetic moment ($M$) of a magnetic dipole can also be expressed as the product of its pole strength ($m$) and magnetic length ($L$):

$M=m\&\#x22C5;L$

Rearranging for pole strength:

$m=\frac{M}{L}$

Using the unit of magnetic moment derived above ($J{T}^{\&\#x2212;1}$) and length in meters ($m$), the unit of pole strength becomes:

$\text{Unit of Pole Strength}=\frac{J{T}^{\&\#x2212;1}}{m}=J{T}^{\&\#x2212;1}{m}^{\&\#x2212;1}$

This matches with (III) in List-II. Thus, (C) matches with (III).

(D) Permeability of free space (${\mathrm{\&\#x03BC;}}_0$):
According to Ampere's Law, the magnetic field $B$ at a distance $r$ from an infinitely long straight current-carrying wire is:

$B=\frac{{\mathrm{\&\#x03BC;}}_{0}I}{2\mathrm{\&\#x03C0;}r}$

Rearranging for ${\mathrm{\&\#x03BC;}}_0$:

${\mathrm{\&\#x03BC;}}_0=\frac{B\&\#x22C5;\left(2\mathrm{\&\#x03C0;}r\right)}{I}$

Substituting the units of magnetic field (Tesla, $T$), distance (meters, $m$), and current (Amperes, $A$):

$\text{Unit of}{\mathrm{\&\#x03BC;}}_0=\frac{T\&\#x22C5;m}{A}=Tm{A}^{\&\#x2212;1}$

This matches with (II) in List-II. Thus, (D) matches with (II).