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

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

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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.
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