#### 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 = [ML^{2}T^{-2}]

Dimensions of charge = [IT]

Thus, the dimensions of voltage can be written as:

[ML^{2}T^{-2}] ×[IT]^{−1} = [ML^{2}I^{−1}T^{−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^{−1}T^{−2}

∴ Dimensions of *vBl* = [LT^{−1}] × [MI^{−1}T^{−2}] × [L]= [ML^{2}I^{−1}T^{−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 = [ML^{2}T^{-2}]

Dimensions of charge = [IT]

The dimensions of voltage can be written as:

[ML^{2}T^{-2}] ×[IT]^{−1} = [ML^{2}I^{−1}T^{−3}]

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