Question

The unit of measurement for magnetic dipole moment of a body is:

• A. A m$^2$
• B. A m$^{-1}$
• C. Wb m$^{-2}$
• D. Wb m$^2$

Hint:

Magnetic dipole moment is always defined as current $\times$ loop area. Keep in mind the distinction: - $H$ (field strength): A·m$^{-1}$, - $B$ (flux density): Wb·m$^{-2}$ = Tesla, - Magnetic moment: A·m$^2$.

Answer 1

The Correct Option is A

Step 1: Recall definition of magnetic dipole moment.
The magnetic dipole moment of a current loop is defined as \[ m = I \cdot A, \] where: - $I$ is the current flowing in the loop (Ampere), - $A$ is the vector area of the loop (m$^2$). Thus, the magnitude of $m$ depends on how much current flows and the area enclosed by that current.

Step 2: Derive SI units.
\[ [m] = [I] \cdot [A] = \text{Ampere} \times \text{m}^2. \] So the unit is A·m$^2$.

Step 3: Why not other options? - Option (B): A·m$^{-1}$ → This is the unit of magnetic field strength $H$, not dipole moment. - Option (C): Wb·m$^{-2}$ → This is the unit of magnetic flux density $B$ (Tesla), not dipole moment. - Option (D): Wb·m$^2$ → This would correspond to flux multiplied by area, which is not relevant here.

Step 4: Physical meaning.
A magnetic dipole moment represents the strength of a magnetic source. A higher value means a stronger tendency to align with an external magnetic field. Final Answer:
\[ \boxed{\text{A m}^2} \]