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The magnetic field perpendicular to the ...

The magnetic field perpendicular to the plane of a conducting ring of radius r changes at the rate.`(dB)/(dt)` . Then

A

the emf induced in the ring is `pi r^(2) (dB)/(dt)`

B

the emf induced in the ring is `2pi r (dB)/(dt)`

C

diametrically opposite points on the ring are at the same potential

D

all points on the ring are at the same potential

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the situation where the magnetic field perpendicular to the plane of a conducting ring is changing over time. We will use Faraday's law of electromagnetic induction to find the induced electromotive force (emf) in the ring. ### Step-by-Step Solution: 1. **Identify the Magnetic Flux (Φ):** The magnetic flux (Φ) through the conducting ring can be expressed as: \[ \Phi = B \cdot A \] where \(B\) is the magnetic field and \(A\) is the area of the ring. For a ring of radius \(r\), the area \(A\) is given by: \[ A = \pi r^2 \] Thus, the magnetic flux becomes: \[ \Phi = B \cdot \pi r^2 \] 2. **Differentiate the Magnetic Flux with Respect to Time:** To find the induced emf (E), we need to differentiate the magnetic flux with respect to time: \[ \frac{d\Phi}{dt} = \frac{d}{dt}(\pi r^2 B) \] Since the radius \(r\) is constant, we can simplify this to: \[ \frac{d\Phi}{dt} = \pi r^2 \frac{dB}{dt} \] 3. **Apply Faraday's Law of Electromagnetic Induction:** According to Faraday's law, the induced emf (E) in the ring is equal to the negative rate of change of magnetic flux: \[ E = -\frac{d\Phi}{dt} \] Substituting our expression for \(\frac{d\Phi}{dt}\): \[ E = -\pi r^2 \frac{dB}{dt} \] 4. **Determine the Sign of the Induced emf:** The negative sign indicates the direction of the induced emf according to Lenz's law, but for the magnitude of the emf, we can write: \[ E = \pi r^2 \frac{dB}{dt} \] 5. **Analyze the Options:** Now we can analyze the options provided in the question: - The first option states that the induced emf is \(E = \pi r^2 \frac{dB}{dt}\), which matches our derived expression. - The second option states \(E = 2\pi r \frac{dB}{dt}\), which is incorrect. - The third option claims that diametrically opposite points on the ring are at the same potential, which is incorrect as there will be a potential difference due to the induced emf. - The fourth option states that all points on the ring are at the same potential, which is correct because the induced emf distributes evenly around the ring. ### Conclusion: The correct answers are: - The induced emf in the ring is given by \(E = \pi r^2 \frac{dB}{dt}\) (Option 1). - All points on the ring are at the same potential (Option 4).
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