Question

In the formula $X = 5Y{Z^2}$, X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y in SI units ?

Answer 1

Hint: We need to find the dimensions of each X and Z and then compare to find the value of Y.

Complete step by step solution:
Given, $X = 5Y{Z^2}$
Taking Y in L.H.S and others on R.H.S we get,
$Y = \dfrac{X}{{5{Z^2}}}$
Now take the dimension of both the sides,
$\left[ Y \right] = \dfrac{{\left[ X \right]}}{{\left[ {{Z^2}} \right]}}$
As 5 is a constant, thus it is dimensionless.
Now put the dimension of capacitance in place of X and put the dimension of magnetic field in place of Y,
$\left[ Y \right] = \dfrac{{{A^2}.{M^{ - 1}}{L^{ - 2}}.{T^4}}}{{{{(M{A^{ - 1}}{T^{ - 2}})}^2}}}$
Simplify the terms we get,
$\left[ Y \right] = {M^{ - 3}}.{L^{ - 2}}.{T^8}.{A^4}$
Thus the dimensions of Y in SI units is ${M^{ - 3}}.{L^{ - 2}}.{T^8}.{A^4}$

Additional information:
The generalised equation for the capacitance of a parallel plate capacitor is given by: C = $\varepsilon \left( {\dfrac{A}{d}} \right)$where $\varepsilon $is the absolute permittivity of the dielectric material being used.
The formula of capacitance can also be given by $q = CV$
A magnetic field is a vector field that describes the magnetic influence on an electric charge of other moving charges. A charge that is moving in a magnetic field experiences a force perpendicular to its own velocity and to the magnetic field.
It is given by$F = qVB$ where F is the Lorentz force, V is the voltage and B is the magnetic field .

Note: A student needs to remember the formula and SI units of capacitance and magnetic field to solve the question.