Chapter 28, Problem 83P

To determine

The angle with the equilibrium position will vary with time according to ${\theta }_0{e}^{-t/2\pi }\cos {\omega }^{\prime }t$ .

Answer to Problem 83P

The angle with the equilibrium will vary according to relation ${\theta }_0{e}^{-t/2\pi }\cos {\omega }^{\prime }t$ .

Explanation of Solution

Given data:

The uniform horizontal magnetic field is $\overrightarrow{B}$ .

The angle $\theta$ is $0$ .

The value of $\tau$ is $RI/{\left(NBA\right)}^2$ .

The value of $\omega$ is $\sqrt{\frac{\kappa }{I}}$ .

The value of ${\omega }^{\prime }$ is ${\omega }_0\sqrt{1-2{\left({\omega }_0\tau \right)}^{-2}}$ .

Formula:

The expression for the restoring torque on the coil is given by,

  ${\tau }_{\text{restoring}}=-\kappa \theta$

The expression for the retarding torque is given by,

  ${\tau }_{\text{restarding}}=NiBA\cos \theta$

The expression for the net torque on the coil is expressed as,

  ${\displaystyle \sum \tau }=I\alpha$

The expression for the induced emf in the coil is evaluated as,

  $\begin{array}{c}\epsilon =-N\frac{d}{dt}\left(BA\sin \theta \right)\\ =-NBA\cos \theta \frac{d\theta }{dt}\end{array}$

The expression for the induced current in the coil is calculated as,

  $i=\frac{\left|\epsilon \right|}{R}$

Calculation:

The net torque acting on the coil is calculated as,

  $\begin{array}{l}{\displaystyle \sum \tau }=I\alpha \\ {\tau }_{\text{restoring}}-{\tau }_{\text{restarding}}=I\frac{d^2\theta }{d{t}^2}\\ -\kappa \theta -NiBA\cos \theta =I\frac{d^2\theta }{d{t}^2}\end{array}$

The magnitude of the current through the coil is calculated as,

  $\begin{array}{c}i=\frac{\left|\epsilon \right|}{R}\\ =\frac{\left|-NBA\cos \theta \frac{d\theta }{dt}\right|}{R}\\ =\frac{NBA\cos \theta \frac{d\theta }{dt}}{R}\end{array}$

The expression for the net torque acting on the torque is further evaluated as,

  $\begin{array}{c}i=\frac{\left|\epsilon \right|}{R}\\ =\frac{\left|-NBA\cos \theta \frac{d\theta }{dt}\right|}{R}\\ =\frac{NBA\cos \theta \frac{d\theta }{dt}}{R}\end{array}$

The expression for the differential is given by,

  $\begin{array}{l}-\kappa \theta -NiBA\cos \theta =I\frac{d^2\theta }{d{t}^2}\\ -\kappa \theta -N\left(\frac{NBA\cos \theta \frac{d\theta }{dt}}{R}\right)BA\cos \theta =I\frac{d^2\theta }{d{t}^2}\\ -\frac{\kappa \theta }{I}-\frac{{\left(NBA\right)}^2}{RI}\frac{d^2\theta }{d{t}^2}=I\frac{d^2\theta }{d{t}^2}\end{array}$

The above equation is evaluated by $\frac{\kappa }{I}$ for ${\omega }_0^2$ and $\frac{{\left(NBA\right)}^2}{RI}$ for $\frac{1}{\tau }$ in the above equation,

  $\frac{d^2\theta }{d{t}^2}+\frac{1}{\tau }\frac{d\theta }{dt}+{\omega }_0^2=0$ …… (I)

The solution for the above differential equation is given by,

  $\theta \left(t\right)={\theta }_0{e}^{-t/2\pi }\cos {\omega }^{\prime }t$

The differential equation $\frac{d\theta }{dt}$ is evaluated as,

  $\begin{array}{l}\frac{d\theta }{dt}=\frac{d\left[{\theta }_0{e}^{-t/2\pi }\cos {\omega }^{\prime }t\right]}{dt}\\ \frac{d\theta }{dt}=-{\theta }_0{e}^{-t/2\pi }\left[{\omega }^{\prime }\sin {\omega }^{\prime }t+\frac{\cos {\omega }^{\prime }t}{2\tau }\right]\end{array}$ ’

The expression for $\frac{d^2\theta }{d{t}^2}$ is evaluated as,

  $\begin{array}{c}\frac{d^2\theta }{d{t}^2}=\frac{d}{dt}\left(-{\theta }_0{e}^{-t/2\pi }\left[{\omega }^{\prime }\sin {\omega }^{\prime }t+\frac{\cos {\omega }^{\prime }t}{2\tau }\right]\right)\\ =\frac{d}{dt}\left(-{\theta }_0{e}^{-t/2\pi }\left[{\omega }^{\prime }\sin {\omega }^{\prime }t+\frac{\cos {\omega }^{\prime }t}{2\tau }\right]\right)\\ =\frac{d}{dt}\left(-{\theta }_0{e}^{-t/2\pi }\left[{\omega }^{\prime }\sin {\omega }^{\prime }t+\frac{\cos {\omega }^{\prime }t}{2\tau }\right]\right)\\ =-{\theta }_0{e}^{-t/2\pi }\left({\omega }^{\prime }\cos {\omega }^{\prime }t-\frac{{\omega }^{\prime }\sin {\omega }^{\prime }t}{2\tau }\right)\\ =-{\theta }_0{e}^{-t/2\pi }\left({{\omega }^{\prime }}^2\cos {\omega }^{\prime }t-\frac{{\omega }^{\prime }}{\tau }\sin {\omega }^{\prime }t+\frac{{\omega }^{\prime }\cos {\omega }^{\prime }t}{4{\tau }^2}\right)\end{array}$

Solve further as,

  $\frac{d^2\theta }{d{t}^2}=-{\theta }_0{e}^{-t/2\pi }\left({{\omega }^{\prime }}^2\cos {\omega }^{\prime }t-\frac{{\omega }^{\prime }}{\tau }\sin {\omega }^{\prime }t+\frac{{\omega }^{\prime }\cos {\omega }^{\prime }t}{4{\tau }^2}\right)$

The values of $\frac{d^2\theta }{d{t}^2}$ and $\frac{d\theta }{dt}$ in equation (I) are evaluated as,

  $\begin{array}{l}\left[\begin{array}{l}-{\theta }_0{e}^{-t/2\pi }\left({{\omega }^{\prime }}^2\cos {\omega }^{\prime }t-\frac{{\omega }^{\prime }}{\tau }\sin {\omega }^{\prime }t+\frac{{\omega }^{\prime }\cos {\omega }^{\prime }t}{4{\tau }^2}\right)\\ +\frac{1}{\tau }\left(-{\theta }_0{e}^{-t/2\pi }\left[{\omega }^{\prime }\sin {\omega }^{\prime }t+\frac{\cos {\omega }^{\prime }t}{2\tau }\right]\right)+{\omega }_0^2\end{array}\right]=0\\ -{\theta }_0{e}^{-t/2\pi }\left({{\omega }^{\prime }}^2\cos {\omega }^{\prime }t+\frac{{\omega }^{\prime }}{4\tau }\cos {\omega }^{\prime }t-{\omega }_0\cos {\omega }^{\prime }t\right)=0\\ \cos {\omega }^{\prime }t\left({{\omega }^{\prime }}^2+\frac{1}{4\tau }-{\omega }_0\right)=0\\ {\omega }^{\prime }={\omega }_0\sqrt{\left(1-\frac{1}{{\left(2\tau {\omega }_0\right)}^2}\right)}\end{array}$

Conclusion:

Therefore, the angle with the equilibrium will vary according to relation ${\theta }_0{e}^{-t/2\pi }\cos {\omega }^{\prime }t$ .