Question

The dimensional formula of magnetic flux is:
$A.\quad [M{{L}^{2}}{{T}^{-2}}{{A}^{-1}}]$
$B.\quad [M{{L}^{3}}{{T}^{-2}}{{A}^{-2}}]$
$C.\quad [{{M}^{0}}{{L}^{2}}{{T}^{-2}}{{A}^{-2}}]$
$D.\quad [M{{L}^{2}}{{T}^{-1}}{{A}^{2}}]$

Answer 1

Hint: To find the magnetic flux $({{\phi }_{B}})$, we will use the definition of flux. Flux refers to the amount of lines of force passing through a Area (A). Hence, the magnetic flux is given by: ${{\phi }_{B}}=B.A$, where (B) refers to the magnetic field, whose lines of force pass through the area (A). By finding the dimensional values of B and A, we will find the dimensional formula of the magnetic flux.

Complete step-by-step answer:
Let’s start by understanding what flux means. In electromagnetism and vector calculus, the flux is a scalar quantity, which can be found by surface integral of the normal component of the vector field over a surface. In a generic sense, the amount of any vector quantity that flows through a cross sectional area at any point in time is known as the flux of that vector quantity.

Hence, the magnetic flux is the magnetic lines of force of magnitude (B) passing through a surface area (A). \[{{\phi }_{B}}=\mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc}
 \overline{B}.d\overline{A}\Rightarrow \ {{\phi }_{B}}=BA\cos \theta \to (1)\], where \[\theta \] is the angle in between the magnetic lines of force and the Area vector.
To find the value of B, we will consider, the magnetic force formula: ${{F}_{B}}=q(\overline{v}\times \overline{B})\Rightarrow {{F}_{B}}=qvB\sin \theta \to (2)$. Lastly, any force can be given by, $F=ma\to (3)$. Equation (2) is equal to Equation (3) dimensionally.
Hence, we will find B to be: $ma=qvB\sin \theta \Rightarrow B=\dfrac{ma}{qv\sin \theta }$. We must remember that the angle doesn’t have any dimension. And current (I) is given by: \[I=\dfrac{dq}{dt}\Rightarrow It=Q\].

Therefore, the dimensional formula of B would be: \[B=\dfrac{ma}{qv\sin \theta }=\dfrac{ma}{Itv\sin \theta }\Rightarrow [B]=\dfrac{[M][L{{T}^{-2}}]}{[A][T][L{{T}^{-1}}]}=\dfrac{[M{{T}^{-2}}]}{[A]}\Rightarrow [B]=[M{{T}^{-2}}{{A}^{-1}}]\].

Using this dimensional formula of B in equation (1), we get: \[{{\phi }_{B}}=BA\cos \theta \Rightarrow [{{\phi }_{B}}]=[B][A]\Rightarrow [{{\phi }_{B}}]=[M{{T}^{-2}}{{A}^{-1}}][{{L}^{2}}]\].

Therefore the dimensional formula of the magnetic flux is: \[{{\phi }_{B}}=[M{{L}^{2}}{{T}^{-2}}{{A}^{-1}}]\], which is Option A.

Note: It’s important to consider whenever finding the flux, is the angle between the vector quantity and the area vector, when the vector quantity passes through the surface area. In the above case, where we are trying to find the magnetic flux$({{\phi }_{B}})$ we see that, \[{{\phi }_{B}}=\mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc}
 \overline{B}.d\overline{A}\Rightarrow \ {{\phi }_{B}}=BA\cos \theta \]. The angle \[\theta \] is the angle in between the magnetic lines of force and the Area vector. The Area vector is always outward normal, hence perpendicular to the plane of the Surface. Hence, if the surface is parallel to the magnetic lines of force, then angle between the magnetic lines of force and the area vector is, \[\theta ={{90}^{0}}\]. Hence, the magnetic flux will be zero in this case. Similarly, the magnetic flux will be maximum when the surface is perpendicular to the magnetic lines of force.