Question

The rod opq is rotating with angular speed w about point o . A uniform magnetic field b exists perpendicular to the plane.

Answer 1

Answer:

the road OPD is rotating with the angular speed w and the point oven for magnetic field b is perpendicular to the plane

Answer 2

Answer:

The two EMFs E1 and E2 are generated in equal and opposite direction, leading to a net ZERO potential difference across the ends.

Explanation:

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

• In electromagnetism and electronics, the electromotive force is the electrical activity 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 the 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,
• The formula used to calculate the induced EMF of a rotating rod is measured by, $\left(\mathrm{E}=\frac{B l^{2} \omega}{2}\right) .$
• 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, $E=\omega(\vec{R} \times \vec{B})$
• Now the two halves of the rod equal, $O A=O B$
• 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 E1 and E2 are generated in equal and opposite direction, leading to a net ZERO potential difference across the ends.