A circular conducting ring of radius rotates with constant angular velocity Omega about its diameter place along the x-axis or uniform magnetic field is applied along by excess if at time t = 0 the ring is entirely in xy plane the EMF induced in the rings?

Question

Answer

Induction in a Rotating Conducting Ring
When a conducting ring rotates in a magnetic field, an electromotive force (EMF) is induced due to the motion of the charges within the ring. Let's analyze the scenario:

Setup Description
- A circular conducting ring of radius $ r $ rotates about its diameter (x-axis) with a constant angular velocity $ \Omega $.
- A uniform magnetic field $ \vec{B} $ is applied along the z-axis.
- At time $ t = 0 $, the ring lies entirely in the xy-plane.

Magnetic Flux and Rotation
- The magnetic flux $ \Phi $ through the ring is given by:
$$
\Phi = B \cdot A
$$
where $ A $ is the area of the ring. For a circular ring, $ A = \pi r^2 $.
- As the ring rotates, the area vector $ \vec{A} $ changes its orientation, leading to a variation in magnetic flux.

Induced EMF Calculation
- According to Faraday's law of electromagnetic induction, the induced EMF $ \mathcal{E} $ in the ring is given by the rate of change of magnetic flux:
$$
\mathcal{E} = -\frac{d\Phi}{dt}
$$
- The area vector $ \vec{A} $ can be expressed as $ A \cos(\theta(t)) $, where $ \theta(t) = \Omega t $. Thus:
$$
\Phi(t) = B \cdot \pi r^2 \cos(\Omega t)
$$
- Taking the derivative:
$$
\frac{d\Phi}{dt} = -B \cdot \pi r^2 \Omega \sin(\Omega t)
$$
- Therefore, the induced EMF becomes:
$$
\mathcal{E} = B \cdot \pi r^2 \Omega \sin(\Omega t)
$$

Conclusion
The induced EMF in the rotating conducting ring is sinusoidal, with a maximum value of $ B \cdot \pi r^2 \Omega $ occurring when $ \sin(\Omega t) = 1 $. This phenomenon exemplifies the principles of electromagnetic induction in a rotating system within a magnetic field.

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