A metallic rod of length 3.0 ~m is hinged at one end and rotated at...

The motional emf $\varepsilon$ induced across the ends of a rod of length $L$ rotating with angular velocity $\omega$ in a perpendicular magnetic field $B$ is given by:
$\varepsilon = \dfrac{1}{2} B \omega L^2$

Rearranging the formula to solve for the magnetic field $B$, we get:
$B = \dfrac{2\varepsilon}{\omega L^2}$

Given values are:
$\varepsilon = 90 \mathrm{~V}$
$\omega = 100 \mathrm{~rad/s}$
$L = 3.0 \mathrm{~m}$

Substituting these values into the rearranged equation:
$B = \dfrac{2 \times 90}{100 \times (3.0)^2}$
$B = \dfrac{180}{100 \times 9}$
$B = \dfrac{180}{900} = 0.2 \mathrm{~T}$

If the factor of $\dfrac{1}{2}$ is forgotten, the result would incorrectly be $0.1 \mathrm{~T}$. If the length is not squared, the result would incorrectly be $0.6 \mathrm{~T}$.

Answer: $0.2 \mathrm{~T}$