A 2cm diameter ring is moved out of uniform magnetic field of 10T in 0.1s. What is magnitude of induce emf in the ring if normal to ring is parallel to magnetic field at all the time?
Question
A 2cm diameter ring is moved out of uniform magnetic field of 10T in 0.1s. What is magnitude of induce emf in the ring if normal to ring is parallel to magnetic field at all the time?
Solution
The induced emf in a loop due to a changing magnetic field is given by Faraday's law of electromagnetic induction, which states that the induced emf is equal to the rate of change of magnetic flux.
The magnetic flux (Φ) through a loop is given by the product of the magnetic field (B), the area of the loop (A), and the cosine of the angle (θ) between the magnetic field and the normal to the loop. In this case, since the normal to the ring is parallel to the magnetic field at all times, θ = 0 and cos(θ) = 1.
So, the magnetic flux is Φ = B * A.
The area of the ring can be calculated using the formula for the area of a circle, A = πr², where r is the radius of the ring. Given that the diameter of the ring is 2 cm, the radius is 1 cm = 0.01 m.
So, A = π * (0.01 m)² = 0.000314 m².
Therefore, the initial magnetic flux is Φ_initial = B * A = 10 T * 0.000314 m² = 0.00314 Tm².
When the ring is moved out of the magnetic field, the final magnetic flux is Φ_final = 0 Tm².
The change in magnetic flux is ΔΦ = Φ_final - Φ_initial = 0 Tm² - 0.00314 Tm² = -0.00314 Tm².
The rate of change of magnetic flux is ΔΦ/Δt = -0.00314 Tm² / 0.1 s = -0.0314 Tm²/s.
Finally, the magnitude of the induced emf is equal to the rate of change of magnetic flux, so the induced emf is 0.0314 V.
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