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Question: The dimensional formula for magnetic moment of a magnet is: A. \(\left[ {{M^0}{L^2}{T^0}{A^1}} \ri...

The dimensional formula for magnetic moment of a magnet is:
A. [M0L2T0A1]\left[ {{M^0}{L^2}{T^0}{A^1}} \right]
B. [M0L2T0A1]\left[ {{M^0}{L^2}{T^0}{A^{ - 1}}} \right]
C. [M0L2T0A1]\left[ {{M^0}{L^{ - 2}}{T^0}{A^{ - 1}}} \right]
D. [M1L2T0A1]\left[ {{M^1}{L^{ - 2}}{T^0}{A^{ - 1}}} \right]

Explanation

Solution

The magnetic moment of a magnet is a quantity that determines the torques, it will experience in an external magnetic field. A loop of electric current, a bar magnet, on an electron revolving around a nucleus, a molecule and a plant all have magnetic moments.

Complete step by step solution:
he magnetic moment for a current carrying loop is given as M=IA......(i)M = IA......\left( i \right)
Taking of with the side of equation (i)
Unit of M==unit of (IA)
Or, unit of M ==unit of (I) ×\timesunit of (A)
== ampere×metre2ampere \times \,metr{e^2}
Putting the dimension of these units,
Dimension of M=A×L2M = A \times {L^2}

Therefore, option A is correct.

Note:

Fundamental physical quantitiesFundamental units of these physical quantitiesDimension of these fundamental quantities
MassKilogram[M]\left[ M \right]
LengthMeter[L]\left[ L \right]
TimeSecond[T]\left[ T \right]
Electric currentAmpere[A]\left[ A \right]
TemperatureKelvin[θ]\left[ \theta \right]
Luminous intensityCandela[cd]\left[ {cd} \right]
Amount of substanceMole[mol]\left[ {mol} \right]