Physics - EMI/AC Question with Solution | TestHub

When the rod rotates at a constant angular velocity $\omega$, the net torque on it is zero. Thus, the external torque balances the magnetic torque: ${\tau }_{ext}={\tau }_{magnetic}$. The induced EMF across the rod is $E=\frac{1}{2}B{L}^2\omega$. The current in the circuit is $I=E/R=\frac{B{L}^2\omega }{2R}$. The magnetic torque is ${\tau }_{magnetic}={\int }_0^L(idr)Br=IB{\int }_0^{L}rdr=IB\frac{L^2}{2}$. Substituting $I$, we get ${\tau }_{magnetic}=\left(\frac{B{L}^2\omega }{2R}\right)B\frac{L^2}{2}=\frac{B^2{L}^4\omega }{4R}$. Equating ${\tau }_{ext}={\tau }_{magnetic}$, we have ${\tau }_{ext}=\frac{B^2{L}^4\omega }{4R}$, which gives $\omega =\frac{4R{\tau }_{ext}}{B^2{L}^4}$. The power dissipated in the resistor is $P=I^{2}R=(E/R{)}^{2}R=E^2/R=\frac{(B{L}^2\omega /2{)}^2}{R}=\frac{B^2{L}^4{\omega }^2}{4R}$. Substituting the expression for $\omega$: $P=\frac{B^2{L}^4}{4R}{\left(\frac{4R{\tau }_{ext}}{B^2{L}^4}\right)}^2=\frac{B^2{L}^4}{4R}\frac{16R^2{\tau }_{ext}^2}{B^4{L}^8}=\frac{4R{\tau }_{ext}^2}{B^2{L}^4}$. Plugging in the given values: $P=\frac{4\times 2.0\times (0.02{)}^2}{(0.5{)}^2\times (1.0{)}^4}=\frac{8\times 0.0004}{0.25\times 1}=\frac{0.0032}{0.25}=0.0128\text{W}$. The mass of the rod is not needed for this calculation.