Share

# Calculate the Dimensions of (A) ∫ → E . D → L , (B) Vbl and (C) D φ B D T . the Symbols Have Their Usual Meaning. - Physics

ConceptElectromagnetic Induction Questions

#### Question

Calculate the dimensions of (a) $\int \overrightarrow{E} . d \overrightarrow{l,}$ (b) vBl and (c) $\frac{d \Phi_B}{dt}.$ The symbols have their usual meaning.

#### Solution

(a) The quantity $\int \overrightarrow{E} . d \overrightarrow{l,}$ can also be written as :-

$\int \overrightarrow{E} . d \overrightarrow{l,}=V......\text(V = Voltage)$

Unit of voltage is J/C.

Voltage can be written as:-

"Voltage"="Energy"/"Charge"

Dimensions of energy = [ML2T-2]
Dimensions of charge = [IT]
Thus, the dimensions of voltage can be written as:
[ML2T-2] ×[IT]−1 = [ML2I−1T−3]

(b) The quantity vBl is the product of quantities v, B and L.

Dimensions of velocity v = [LT−1]
Dimensions of length l = [L]
The dimensions of magnetic field B can be found using the following formula:-

B=F/(qv)

Dimensions of force F = [MLT−2]
Dimensions of charge q = [IT]
Dimensions of velocity = [LT−1]
The dimensions of a magnetic field can be written as:
MI−1T−2
∴ Dimensions of vBl = [LT−1] × [MI−1T−2] × [L]= [ML2I−1T−3]

(c) The quantity $\frac{d\phi}{dt}$ is equal to the emf induced; thus, its dimensions are the same as that of the voltage.
Voltage can be written as:-

"Voltage"="Energy"/"Charge"

Dimensions of energy = [ML2T-2]
Dimensions of charge = [IT]
The dimensions of voltage can be written as:
[ML2T-2] ×[IT]−1 = [ML2I−1T−3]
∴ Dimensions of $\frac{d\phi}{dt}=ML^2I^{−1}T^{−3}$

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [1]

Solution Calculate the Dimensions of (A) ∫ → E . D → L , (B) Vbl and (C) D φ B D T . the Symbols Have Their Usual Meaning. Concept: Electromagnetic Induction Questions.
S