Question

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

Answer 1

(a) The quantity$\int$E.dl can also be written as:
$\int$E.dl = V (V = Voltage)
Unit of voltage is J/C.
Voltage can be written as:
$\mathrm{Voltage}=\frac{\mathrm{Energy}}{\mathrm{Charge}}$
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:
$B=\frac{F}{qv}$
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 $\frac{d\mathrm{\varphi }}{dt}$ is equal to the emf induced; thus, its dimensions are the same as that of the voltage.
Voltage can be written as:
$\mathrm{Voltage}=\frac{\mathrm{Energy}}{\mathrm{Charge}}$
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 $\frac{d\mathrm{\varphi }}{dt}$ = [ML²I⁻¹T⁻³]