A metal disc of radius $r$ can rotate with negligible friction inside a long, straight coil, about a shaft parallel to the axis of symmetry of the coil. One end of the coil wire is connected to the edge of the disc and the other to the shaft. 

The coil has resistance $R$ and contains n turns per unit length. It is placed so that its axis is parallel to the Earth’s magnetic field vector ${\overrightarrow{B}}_0$. What current flows through the ammeter shown in Fig., if the disc rotates with angular frequency $\omega$? Plot the current as a function $\omega$ for both directions of rotation. If the power developed in the ohmic coil is P_el and if the mechanical power delivered is P_mech then

The resultant magnetic field $B\text{'}$ is the sum of magnetic fields of the Earth and the coil $B_0$ and $B$ respectively.

i.e., $B\text{'}=B_0\pm B\dots \dots \dots \dots \dots \left(1\right)$

The current flowing through the coil is 

$I=\frac{V_i}{R}=\frac{B\text{'}{r}^2\omega }{2R}\dots \dots \dots \dots \dots ..\left(2\right)$

The magnetic field produced by the coil itself is 

$B={\mu }_{0}nI\dots \dots \dots \dots \dots ..\left(3\right)$

From the equations $\left(1\right), \left(2\right)$ and $\left(3\right)$ above, $B,B\text{'}$ and $I$ can be determined. The two signs occurring the equation $\left(1\right)$ allows for both positive and negative values of the angular frequency. The value of \omega is taken to be positive if the magnetic field of the coil acts in the same direction as that of the Earth. The following results are obtained for the resultant magnetic field and the current.

$B\text{'}=\frac{2R{B}_0}{2R-{\mu }_{0}n{r}^2\omega }$ and $I=\frac{B_0{r}^2\omega }{2R-{\mu }_{0}n{r}^2\omega }$

Note that when the disc is at rest, the current is zero and the resultant magnetic field inside the coil is simply $B_0$, the magnetic field of the Earth.

Case (i): When the direction of rotation is such that the field in the coil oppose the external magnetic field. $(\omega <0)$

The resultant magnetic field decreases asymptotically to zero as the (negative) angular frequency of the disc increases. At such high rotation speeds, the current flowing in the coil tends to $–B_0/{\mu }_{0}n$ (the value needed to cancel the magnetic field of the Earth.)

Case (ii): Rotation of the disc in the opposite direction $(\omega >0)$

It causes the resultant magnetic field to increases. This leads to a higher voltage being induced and a larger current flowing which in turn leads to a further increase in the magnetic field. Both the magnetic field and the current tend to infinity as a ‘critical’ angular frequency, ${\omega }_{crit}=2R\left({\mu }_{0}n{r}^2\right)$, is approached as shown in Fig. (b). Such a state is not realized in practice, the current and 

The heat given out by the coil increase until the wire burn away! The relationship between the current in the coil and the resultant magnetic field is represented graphically as in the figure (c). According to equation $\left(2\right)$, $I$ is proportional to $B\text{'}$, the coefficient of proportionality depends on $\omega$.

This is represented by a straight line through the origin, with a slope proportional to $\omega$. Equation $\left(1\right)$ and $\left(3\right)$ show that $B\text{'}=B_0+{\mu }_{0}nI$, this is also a linear relationship but its graph does not pass through the origin.

The gradient of the latter is $1/\left({\mu }_{0}n\right)$ and independent of $\omega$. The intersection of these two straight lines determines the actual current and the resultant magnetic field. If $\omega ={\omega }_{crit}$, then the gradients of the two straight lines are the same and the equations have no solution.

The electrical power is $P_{el}=I^{2}R$, while the mechanical work done per second is the product of the torque and the angular frequency, $P_{mech}=\tau \omega =B\text{'}I{r}^2\omega /2$. The torque $\tau$ is the product of the force $B\text{'}Ir$ and the average perpendicular distance of its line of action from the axis, $r/2.$

$P_{el}=I^{2}R={\left[\frac{B_0{r}^2\omega }{2R-{\mu }_{0}n{r}^2\omega }\right]}^{2}R$

$P_{mech}=\frac{1}{2}B\text{'}I{r}^2\omega =\frac{1}{2}\left[\frac{2B_{0}R}{2R-{\mu }_{0}n{r}^2\omega }\right]{\left[\frac{B_0{r}^2\omega }{2R-{\mu }_{0}n{r}^2\omega }\right]}^2{r}^2\omega$

           $={\left[\frac{B_0{r}^2\omega }{2R-{\mu }_{0}n{r}^2\omega }\right]}^{2}R$

Thus, $P_{el}=P_{mech}$