Question

A metallic rod of 1.0 m length is rotating about a perpendicular axis passing through its one end with an angular frequency of 400 rad/s. The other end of the rod is in contact with a ring of metal. Magnetic field of 0.5 T is along its axis. Calculate the induced emf between the ring and the axis.

Hint:

A common mistake is to use the linear velocity of the end of the rod (\(v = \omega L\)) and the formula for a linearly moving rod (\(\mathcal{E} = BvL\)). This gives \(\mathcal{E} = B(\omega L)L = B\omega L^2\), which is incorrect by a factor of 2. Always remember the factor of \(\frac{1}{2}\) for a rotating rod, which comes from the integration.

Answer 1

Step 1: Understanding the Concept (Motional EMF):
When a conducting rod rotates in a uniform magnetic field, an electromotive force (EMF) is induced across its ends. This is a type of motional EMF. Each point on the rod moves with a different linear speed, so we must integrate the small EMFs generated across infinitesimal segments of the rod to find the total EMF.

Step 2: Key Formula and Derivation:
Consider a small element of the rod of length \(dr\) at a distance \(r\) from the axis of rotation.
The linear velocity of this element is given by \(v = \omega r\), where \(\omega\) is the angular frequency.
The small EMF induced across this element \(dr\) is: \[ d\mathcal{E} = B v dr = B(\omega r) dr \] To find the total EMF induced between the axis (\(r=0\)) and the outer end (the ring at \(r=L\)), we integrate this expression from \(0\) to \(L\): \[ \mathcal{E} = \int_{0}^{L} d\mathcal{E} = \int_{0}^{L} B\omega r \, dr \] Since \(B\) and \(\omega\) are constant: \[ \mathcal{E} = B\omega \int_{0}^{L} r \, dr = B\omega \left[ \frac{r^2}{2} \right]_{0}^{L} = B\omega \left( \frac{L^2}{2} - 0 \right) \] \[ \mathcal{E} = \frac{1}{2} B \omega L^2 \]

Step 3: Calculation:
We are given the following values: \begin{itemize} \item Length of the rod, \(L = 1.0\) m. \item Angular frequency, \(\omega = 400\) rad/s. \item Magnetic field strength, \(B = 0.5\) T. \end{itemize} Substitute these values into the derived formula: \[ \mathcal{E} = \frac{1}{2} (0.5 \text{ T}) (400 \text{ rad/s}) (1.0 \text{ m})^2 \] \[ \mathcal{E} = \frac{1}{2} \times 0.5 \times 400 \times 1 \] \[ \mathcal{E} = 0.25 \times 400 \] \[ \mathcal{E} = 100 \text{ V} \]

Step 4: Final Answer:
The induced emf between the ring and the axis is 100 V.