Question

A metallic rod of length l is rotated at an angular speedw , normal to a uniform magnetic field B. Derive expressions for
1) emf induced in the rod
2) heat dissipiation, if the resistance of the rod is R

Answer 1

$$\begin{gathered} \mathbf{Given}\mathbf{}\mathbf{data}\mathbf{:} \\ \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{metal}\mathrm{rod}=\mathrm{L} \\ \mathrm{Angular}\mathrm{velocity}=\mathrm{\omega } \\ \mathrm{Magnetic}\mathrm{field}=\mathrm{B} \\ \mathrm{Resistance}\mathrm{of}\mathrm{the}\mathrm{rod}=\mathrm{R} \\ \mathrm{dx}=\mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{element}\mathrm{at}\mathrm{distance}\mathrm{x}\mathrm{from}\mathrm{the}\mathrm{centre} \\ \mathrm{v}=\mathrm{Linear}\mathrm{velocity} \\ \mathrm{Solution}: \\ \frac{\mathrm{dA}}{\mathrm{dt}}=\mathrm{Area}\mathrm{swept}\mathrm{by}\mathrm{the}\mathrm{element}\mathrm{per}\mathrm{second}=\mathrm{v}\mathrm{dx} \\ \mathrm{Q}1) \\ \mathbf{Induced}\mathbf{}\mathbf{emf}\mathbf{:} \\ \mathrm{dE}=\mathrm{B}\frac{\mathrm{dA}}{\mathrm{dt}} \\ =\mathrm{Bvdx} \\ =\mathrm{B}\left[\mathrm{x\omega }\right]\mathrm{dx} \\ \mathrm{Emf}\mathrm{induced}\mathrm{across}\mathrm{the}\mathrm{rod}: \\ \mathrm{E}={\int }_0^{\mathrm{L}}\mathrm{Bx\omega }d\mathrm{x} \\ ={\left[\frac{{\mathrm{B\omega x}}^2}{2}\right]}_0^{\mathrm{L}} \\ =\frac{1}{2}{\mathrm{B\omega L}}^2 \\ --------------------- \\ \mathrm{Q}2) \\ \mathrm{Current}\mathrm{in}\mathrm{the}\mathrm{rod}: \\ \mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}} \\ =\frac{1}{2}\frac{{\mathrm{B\omega L}}^2}{\mathrm{R}} \\ \mathbf{Heat}\mathbf{}\mathbf{dissipated}\mathbf{:} \\ \mathrm{H}={\mathrm{I}}^2\mathrm{Rt} \\ ={\left[\frac{1}{2}\frac{{\mathrm{B\omega L}}^2}{\mathrm{R}}\right]}^2\times \mathrm{Rt} \\ \mathbf{}\mathbf{=}\frac{{\mathbf{B}}^{\mathbf{2}}{\mathbf{\omega }}^{\mathbf{2}}{\mathbf{L}}^{\mathbf{4}}\mathbf{t}}{\mathbf{4}\mathbf{R}} \end{gathered}$$