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.

Answer 1

(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 \[\frac{d\phi}{dt}=ML^2I^{−1}T^{−3}\]