Question

What is the dimension of magnetic field B in terms of C (Coulomb), M (Mass), L (Length), T (Time)?
$
  {\text{A}}{\text{. }}\left[ {{{\text{M}}^1}{{\text{L}}^1}{{\text{T}}^{ - 2}}{\text{C}}} \right] \\
  {\text{B}}{\text{. }}\left[ {{{\text{M}}^1}{{\text{L}}^0}{{\text{T}}^{ - 1}}{{\text{C}}^{ - 1}}} \right] \\
  {\text{C}}{\text{. }}\left[ {{{\text{M}}^1}{{\text{L}}^0}{{\text{T}}^{ - 2}}{\text{C}}} \right] \\
  {\text{D}}{\text{. }}\left[ {{{\text{M}}^1}{{\text{L}}^0}{{\text{T}}^{ - 1}}{\text{C}}} \right] \\
 $

Answer 1

Hint- Here, we will proceed by defining the term magnetic field. Then, we will be using a magnetic force formula to represent the magnetic field in terms of force, charge, length and time. Then, dimensional analysis is done on formula.
Formulas used: \[{\text{F}} = {\text{ILB}}\sin \theta \] and ${\text{I}} = \dfrac{{\text{q}}}{{\text{t}}}$.

Complete Step-by-Step solution:
Magnetic field is a vector field in the neighbourhood of a magnet, electric current, or changing electric field, in which magnetic forces are observable. Magnetic fields such as that of Earth cause magnetic compass needles and other permanent magnets to line up in the direction of the field. Magnetic fields force moving electrically charged particles in a circular or helical path.
According to Magnetic Force Formula
When a wire carrying electric charge is placed in a magnetic field, a force is exerted on the wire. This force is given by the mentioned formula in vector form.
$\overrightarrow {\text{F}} = {\text{I}}\overrightarrow {\text{L}} \times \overrightarrow {\text{B}} $ where I represents the magnitude of the current passing through the wire, $\overrightarrow {\text{L}} $ represents the length vector and $\overrightarrow {\text{B}} $ represents the magnetic field vector.
If $\theta $ is the angle between the length vector and the magnetic field vector, then the vector form can be represented in scalar form as under
\[
  {\text{F}} = {\text{ILB}}\sin \theta \\
   \Rightarrow {\text{B}} = \dfrac{{\text{F}}}{{{\text{IL}}\sin \theta }}{\text{ }} \to {\text{(1)}} \\
 \]
Since, current is defined as charge per unit time i.e., ${\text{I}} = \dfrac{{\text{q}}}{{\text{t}}}$ where I represents the current flowing, q represents the charge and t represents the time.
By putting ${\text{I}} = \dfrac{{\text{q}}}{{\text{t}}}$ in equation (1), we get
\[
   \Rightarrow {\text{B}} = \dfrac{{\text{F}}}{{\left( {\dfrac{{\text{q}}}{{\text{t}}}} \right){\text{L}}\sin \theta }} \\
   \Rightarrow {\text{B}} = \dfrac{{{\text{Ft}}}}{{{\text{qL}}\sin \theta }}{\text{ }} \to {\text{(2)}} \\
 \]
As we know that the dimension of force (F) is $\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]$, dimension of time (t) is [T], dimension of charge (q) is [C], dimension of length (L) is [L] and \[\sin \theta \] is dimensionless.
By putting all these dimensions in the RHS of equation (2), we get
Dimensions of Magnetic field (B) = \[\dfrac{{\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]\left[ {\text{T}} \right]}}{{\left[ {\text{C}} \right]\left[ {\text{L}} \right]}} = \left[ {{\text{ML}}{{\text{T}}^{ - 2}}{\text{T}}{{\text{L}}^{ - 1}}{{\text{C}}^{ - 1}}} \right] = \left[ {{\text{M}}{{\text{L}}^{1 - 1}}{{\text{T}}^{ - 2 + 1}}{{\text{C}}^{ - 1}}} \right] = \left[ {{{\text{M}}^1}{{\text{L}}^0}{{\text{T}}^{ - 1}}{{\text{C}}^{ - 1}}} \right]\]
Therefore, the dimensions of magnetic field is \[\left[ {{{\text{M}}^1}{{\text{L}}^0}{{\text{T}}^{ - 1}}{{\text{C}}^{ - 1}}} \right]\].
Hence, option B is correct.

Note- The dimension of force (F) is $\left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]$ because force is simply the product of mass (having dimension [M]) and acceleration (having dimension $\left[ {{\text{L}}{{\text{T}}^{ - 2}}} \right]$). So, Dimensions of force = $\left[ {\text{M}} \right]\left[ {{\text{L}}{{\text{T}}^{ - 2}}} \right] = \left[ {{\text{ML}}{{\text{T}}^{ - 2}}} \right]$where acceleration is simply velocity per unit time (where velocity is rate of change of position).