Question

The dimension of mutual inductance is (Denote dimension of current as A)

• A. \( \text{M L}^2 \text{ T}^{-2} \text{ A}^{-2} \)
• B. \( \text{M L}^2 \text{ T}^{-2} \text{ A}^{-2} \)
• C. \( \text{M L}^{-2} \text{ T}^2 \text{ A}^{-2} \)
• D. \( \text{M L}^2 \text{ T}^{-3} \text{ A}^{-1} \)
• E. \( \text{M L}^2 \text{ T}^{-3} \text{ A}^{-3} \)

Hint:

Dimensional analysis is often easiest using energy formulas (\( \frac{1}{2}LI^2 \) or \( \frac{1}{2}CV^2 \)) as energy units are well-known.

Answer 1

The Correct Option is A

Concept: The energy (\( U \)) stored in an inductor is given by \( U = \frac{1}{2} L I^2 \). Mutual inductance \( M \) shares the same dimensions as self-inductance \( L \).

Step 1: {Relate energy to inductance.}
Energy dimensions: \( [U] = [\text{M L}^2 \text{ T}^{-2}] \).
Current dimensions: \( [I] = [\text{A}] \).

Step 2: {Isolate the dimensions of M.}
\[ [M] = \frac{[U]}{[I]^2} = \frac{\text{M L}^2 \text{ T}^{-2}}{\text{A}^2} \] \[ [M] = \text{M L}^2 \text{ T}^{-2} \text{ A}^{-2} \]