The dimensional formula magnetic moment of a magnet is - M - 0 - L - 2 - T - 0 - A - -1 - M - 0 - L - -2 - T - 0 - A - -1 - M - 0 - L - 2 - T - 0 - A - 1 - M - 1 - L - -2 - T - 0 - A - -1 -

Question

Toppr Admin · Accepted Answer

We know that magnetic moment for a solenoid is NIS- where N is number of loops- I is current and S-xA0-is the vector area-Thus- dimensions of -x3BC- are -AL2-M0L2T0A-

List-I	List-II
A.Spring constant	1. $M^{1} L^{2} T^{- 2}$
B.pascal	2. $M^{0} L^{0} T^{- 1}$
C.hertz	3. $M^{1} L^{0} T^{- 2}$
D.joule	4. $M^{1} L^{- 1} T^{- 2}$

List - I	List - II
A) Spring constant	I) [ $M^{1} L^{2} T^{- 2}$ ]
B) Pascal	II) [ $M^{0} L^{0} T^{- 1}$ ]
C) Hertz	III)[ $M^{1} L^{0} T^{- 2}$ ]
D) Joule	IV) [ $M^{1} L^{- 1} T^{- 2}$ ]

The dimensional formula for magnetic moment of a magnet is :
$[M^{0} L^{2} T^{0} A^{- 1}]$
$[M^{0} L^{- 2} T^{0} A^{- 1}]$
$[M^{0} L^{2} T^{0} A^{1}]$
$[M^{1} L^{- 2} T^{0} A^{- 1}]$

We know that magnetic moment for a solenoid is NIS, where N is number of loops, I is current and S is the vector area.
Thus, dimensions of $μ$ are $[A L^{2}] = [M^{0} L^{2} T^{0} A]$

The dimensional formula for magnetic moment of a magnet is :[M0L2T0A−1][M0L−2T0A−1][M0L2T0A1][M1L−2T0A−1]

We know that magnetic moment for a solenoid is NIS, where N is number of loops, I is current and S is the vector area.Thus, dimensions of μ are [AL2]=[M0L2T0A]

The dimensional formula for magnetic moment of a magnet is :
$[M^{0} L^{2} T^{0} A^{- 1}]$
$[M^{0} L^{- 2} T^{0} A^{- 1}]$
$[M^{0} L^{2} T^{0} A^{1}]$
$[M^{1} L^{- 2} T^{0} A^{- 1}]$

We know that magnetic moment for a solenoid is NIS, where N is number of loops, I is current and S is the vector area.
Thus, dimensions of $μ$ are $[A L^{2}] = [M^{0} L^{2} T^{0} A]$