Question

A rod of length l rotates with a uniform angular velocity $\omega$rad/s about an axis passing through its middle point but normal to its length in a uniform magnetic field of induction B with its direction parallel to the axis of rotation. The induced EMF between the two ends of the rod is:
$\left( A \right)\dfrac{{B{l^2}\omega }}{2}$
$\left( B \right)$ Zero
$\left( C \right)\dfrac{{B{l^2}\omega }}{4}$
$\left( D \right)2B{l^2}\omega$

Answer 1

In this particular question use the concept that the induced EMF of a rotating rod is measured by, (${\rm E} = \dfrac{{B{l^2}\omega }}{2}$), where B is flux density, l is length of the rod and $\omega$ is angular velocity of the rod, also use the property that induced EMF, $E = \omega \left( {\vec R \times \vec B} \right)$, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Now,
In electromagnetism and electronics, electromotive force is the electrical action produced by a non-electrical source. Devices (known as transducers) provide an EMF by converting other forms of energy into electrical energy, such as batteries or generators. In electromagnetic induction, EMF can be defined around a closed loop of conductor as the electromagnetic work that would be done on an electric charge if it travels once around the loop. For a time varying magnetic flux linking a loop, the electric potential’s scalar field is not defined due to a circulating electric vector field, but an EMF nevertheless does work that can be measured as a virtual electric potential around the loop. We use the simple formula to calculate the induced EMF of a rotating rod. The formula used to calculate the induced EMF of a rotating rod is measured by, (${\rm E} = \dfrac{{B{l^2}\omega }}{2}$).
Note that an EMF always generated along the direction ($\vec V \times \vec B$)
For circular motion, $\vec V = \vec R\omega$
Hence, EMF induced, $E = \omega \left( {\vec R \times \vec B} \right)$
Now the two halves of the rod equal, OA = OB
Thus, $\left| {{{\vec R}_1}} \right| = \left| {{{\vec R}_2}} \right|$
But, ${\vec R_1} = - {\vec R_2}$
Therefore, the induced EMF will be in the opposite direction from the point O. Thus, the two EMFs ${{\rm E}_1}$ and ${{\rm E}_2}$ are generated in equal and opposite direction, leading to a net ZERO potential difference across the ends.

So, the correct answer is “Option B”.

Note: As we already studied that, what is EMF and how to calculate it for a rotating rod in a magnetic field. So, we used the simple formula for calculating the EMF of a rotating rod in the given situation. We find out that (${\vec R_1} = - {\vec R_2}$), therefore, the net induced EMF in the rod will be zero. This gives us the correct option for the above question.