13.3 Motional emf  (Page 3/6)

A metal rod rotating in a magnetic field

Strategy

Solution

1. From geometry, the area of the loop OPSO is $A=\frac{r^2\theta }{2}.$ Hence, the magnetic flux through the loop is
   
   ${\text{Φ}}_{\text{m}}=BA=B\frac{r^2\theta }{2}.$
   
   Differentiating with respect to time and using $\omega =d\theta \text{/}dt,$ we have
   
   $\epsilon =\left|\frac{d{\text{Φ}}_{\text{m}}}{dt}\right|=\frac{B{r}^2\omega }{2}.$
   
   When divided by the resistance R of the loop, this yields for the magnitude of the induced current
   
   $I=\frac{\epsilon }{R}=\frac{B{r}^2\omega }{2R}.$
   
   As $\theta$ increases, so does the flux through the loop due to $\overrightarrow{B}.$ To counteract this increase, the magnetic field due to the induced current must be directed into the page in the region enclosed by the loop. Therefore, as part (b) of [link] illustrates, the current circulates clockwise.
2. You rotate the rod by exerting a torque on it. Since the rod rotates at constant angular velocity, this torque is equal and opposite to the torque exerted on the current in the rod by the original magnetic field. The magnetic force on the infinitesimal segment of length dx shown in part (c) of [link] is $d{F}_{\text{m}}=IBdx,$ so the magnetic torque on this segment is
   
   $d{\tau }_{\text{m}}=x·d{F}_{\text{m}}=IBxdx.$
   
   The net magnetic torque on the rod is then
   
   ${\tau }_{\text{m}}={\displaystyle {\int }_0^{r}d{\tau }_{\text{m}}}=IB{\displaystyle {\int }_0^{r}xdx=\frac{1}{2}}IB{r}^2.$
   
   The torque $\tau$ that you exert on the rod is equal and opposite to ${\tau }_{\text{m}},$ and the work that you do when the rod rotates through an angle $d\theta$ is $dW=\tau d\theta .$ Hence, the work per unit time that you do on the rod is
   
   $\frac{dW}{dt}=\tau \frac{d\theta }{dt}=\frac{1}{2}IB{r}^2\frac{d\theta }{dt}=\frac{1}{2}\left(\frac{B{r}^2\omega }{2R}\right)B{r}^2\omega =\frac{B^2{r}^4{\omega }^2}{4R},$
   
   where we have substituted for I . The power dissipated in the resister is $P=I^{2}R$ , which can be written as
   
   $P={\left(\frac{B{r}^2\omega }{2R}\right)}^{2}R=\frac{B^2{r}^4{\omega }^2}{4R}.$
   
   Therefore, we see that
   
   $P=\frac{dW}{dt}.$
   
   Hence, the power dissipated in the resistor is equal to the work per unit time done in rotating the rod.