Answer

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**Hint**The electric field $ E $ can be defined as the negative of the gradient of the potential between two points. The potential can also have a unit of Tesla square meter per second. So using this we can find our answer.

Formula used: In this solution we will be using the following formulae;

$ V = Blv $ where $ V $ is potential difference, $ B $ is the magnetic field, $ l $ is the length of an object in that magnetic field and $ v $ is the velocity of the object.

$ E = - \dfrac{{\Delta V}}{L} $ where $ E $ is the electric field, $ \Delta V $ is change in potential and $ L $ is the length across which the potential is measured.

**Complete step by step answer**

The dimension of any variable can be written as the combination of all the dimensions of the variables which constitute that variable with their respective operation, this is to say, for example, that the variable $ I $ which is current and its fundamental unit is $ {\text{A}} $ which is essentially coulomb per second, can be written as $ {\text{V/}}\Omega $ because the current can be expressed as Voltage over resistance as in $ I = \dfrac{V}{R} $

Similarly, the potential difference or voltage across a particular length in a magnetic field can be given as; $ V = Blv $ where $ B $ is the magnetic field, $ l $ is the length of an object in that magnetic field and $ v $ is the velocity of the object, hence its dimension can be written as the product of the dimension of all the variables as in

$ \left[ V \right] = \left[ B \right]\left[ L \right]\left[ v \right] $ where the square bracket signifies dimension.

Now $ \left[ v \right] $ can be written as $ \left[ L \right]{\left[ T \right]^{ - 1}} $ since velocity is distance per unit time. Hence, by replacement

$ \left[ V \right] = \left[ B \right]\left[ L \right]\left[ L \right]{\left[ T \right]^{ - 1}} $

Now, $ E = - \dfrac{{\Delta V}}{L} $ where $ E $ is the electric field, $ \Delta V $ is change in potential and $ L $ is the length across which the potential is measured. Hence

$ \left[ E \right] = \left[ V \right]{\left[ L \right]^{ - 1}} $ then $ \left[ V \right] = \left[ E \right]\left[ L \right] $

Substituting into $ \left[ V \right] = \left[ B \right]\left[ L \right]\left[ L \right]{\left[ T \right]^{ - 1}} $ we have

$ \left[ E \right]\left[ L \right] = \left[ B \right]\left[ L \right]\left[ L \right]{\left[ T \right]^{ - 1}} $

By cancelling common terms we have

$ \left[ E \right] = \left[ B \right]\left[ L \right]{\left[ T \right]^{ - 1}} $

**Hence, the correct answer is option C.**

**Note**

Alternatively, we can deal with units, and then convert to its proper dimension as in

$ \left( {N/C} \right) = V/m $ and $ {\text{V}} = {\text{T}}{m^2}{s^{ - 1}} $ where $ {\text{T}} $ is tesla not time.

$ V = \left( {N/C} \right)m = {\text{T}}{m^2}{s^{ - 1}} $

Hence by cancellation, we have

$ \left( {N/C} \right) = {\text{T}}m{s^{ - 1}} $ which is

$ \left[ E \right] = \left[ B \right]\left[ L \right]{\left[ T \right]^{ - 1}} $

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