Question

The dimensions of the coefficient of self-inductance are:

• A. \( [M L^2 T^{-2} A^{-2}] \)
• B. \( [M L^2 T^{-2} A^{-1}] \)
• C. \( [M L T^{-2} A^{-2}] \)
• D. \( [M L T^{-2} A^{-1}] \)

Hint:

Self-inductance is the property of an inductor that determines how much EMF is induced per unit rate of change of current. Its dimensional formula is derived using the energy equation.

Answer 1

The Correct Option is A

Step 1: {Define self-inductance} 
The energy stored in an inductor is given by: \[ U = \frac{1}{2} L I^2 \] where \( L \) is the self-inductance and \( I \) is the current. 
Step 2: {Rearrange for \( L \)} 
\[ L = \frac{2U}{I^2} \] 
Step 3: {Find the dimensional formula of \( L \)} 
Since energy (\( U \)) has the dimensional formula: \[ [U] = [M L^2 T^{-2}] \] and current (\( I \)) has the dimensional formula: \[ [I] = [A] \] we substitute: \[ [L] = \frac{[M L^2 T^{-2}]}{[A^2]} \] \[ = [M L^2 T^{-2} A^{-2}] \] 
Step 4: {Verify the options} 
Comparing with the given choices, 
the correct answer is (A) \( [M L^2 T^{-2} A^{-2}] \). 
 

Answer 2

Given:
We are to find the dimensions of the coefficient of self-inductance, denoted by \( L \).

Step 1: Use the formula for self-inductance
The induced EMF is given by:
\[ \mathcal{E} = -L \frac{dI}{dt} \Rightarrow L = \frac{\mathcal{E}}{dI/dt} \]
Where:
- \( \mathcal{E} \) is the electromotive force with dimensions \( [ML^2T^{-3}A^{-1}] \)
- \( \frac{dI}{dt} \) is the rate of change of current with dimensions \( [AT^{-1}] \)

Step 2: Divide the dimensions
\[ [L] = \frac{[ML^2T^{-3}A^{-1}]}{[AT^{-1}]} = [ML^2T^{-2}A^{-2}] \]

Final Answer:
The dimensions of the coefficient of self-inductance are \( \boxed{[ML^2T^{-2}A^{-2}]} \)