Question

The dimensions of the rate of change of magnetic flux are :

• A. \([M\,L\,T^{-3}\,A]\)
• B. \([M\,L^{2}\,T^{-3}\,A^{-1}]\)
• C. \([M\,L^{2}\,T^{-2}\,A^{-1}]\)
• D. \([M\,L^{2}\,T^{-3}\,A^{-2}]\)

Hint:

Remember Faraday's law: \[ \mathcal{E}=-\frac{d\Phi}{dt}. \] Therefore, the dimensions of the rate of change of magnetic flux are always the same as the dimensions of emf: \[ [M\,L^{2}\,T^{-3}\,A^{-1}]. \]

Answer 1

The Correct Option is B

Concept: According to Faraday's law of electromagnetic induction, \[ \mathcal{E} = -\frac{d\Phi}{dt}, \] where \[ \Phi = \text{magnetic flux}. \] Thus, the rate of change of magnetic flux has the same dimensions as emf. Therefore, instead of directly finding dimensions of magnetic flux and differentiating, we can use the dimensions of emf.

Step 1: Write the dimensions of emf. Emf is defined as work done per unit charge. \[ \text{EMF} = \frac{\text{Work}}{\text{Charge}}. \]

Step 2: Find dimensions of work. Work \[ = \text{Force}\times\text{Distance}. \] Dimensions of force are \[ [M\,L\,T^{-2}]. \] Therefore, \[ [\text{Work}] = [M\,L\,T^{-2}] \times [L] = [M\,L^{2}\,T^{-2}]. \]

Step 3: Find dimensions of charge. Charge is \[ Q=It. \] Hence, \[ [Q]=[A\,T]. \]

Step 4: Determine dimensions of emf. Using \[ \text{EMF} = \frac{\text{Work}}{\text{Charge}}, \] we obtain \[ [\text{EMF}] = \frac{[M\,L^{2}\,T^{-2}]}{[A\,T]}. \] Therefore, \[ [\text{EMF}] = [M\,L^{2}\,T^{-3}\,A^{-1}]. \]

Step 5: Apply Faraday's law. Since \[ \mathcal{E} = -\frac{d\Phi}{dt}, \] the dimensions of \[ \frac{d\Phi}{dt} \] are exactly the same as those of emf. Hence, \[ \boxed{ \left[\frac{d\Phi}{dt}\right] = [M\,L^{2}\,T^{-3}\,A^{-1}] }. \]

Step 6: Choose the correct option. Comparing with the given options, \[ \boxed{\text{(B)}} \] is the correct answer.