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Chapter 14: Problem 2013

A ring of radius \(\mathrm{r}\) is rotating about its diameter with angular velocity w in a perpendicular magnetic field \(\mathrm{B}^{-}\) It has 20 turns. The emf induced is (a) \(20 \mathrm{~B} \pi \mathrm{r}^{2} \sin \mathrm{wt}\) (b) \(20 \mathrm{~B} \pi \mathrm{r}^{2} \mathrm{cos} \mathrm{wt}\) (c) \(10 \sqrt{2} \mathrm{~B} \pi \mathrm{r}^{2}\) (d) \(20 \mathrm{~B} \pi \mathrm{r}^{2} \mathrm{w} \sin \mathrm{wt}\)

Short Answer

Expert verified
The induced emf in the rotating ring is given by the expression \(emf = 20 B \pi r^2 w \sin(wt)\), so the correct answer is (a) \(20 B \pi r^{2} \sin(wt)\).

Step by step solution

01

Recall Faraday's law of electromagnetic induction

According to Faraday's law, the induced emf in a loop is proportional to the rate of change of magnetic flux through the loop. Mathematically, it can be expressed as: \(emf = -N \frac{dΦ}{dt}\) where \(emf\) is the induced emf, \(N\) is the number of turns, and \(\frac{dΦ}{dt}\) is the rate of change of magnetic flux.
02

Calculate the magnetic flux through the ring

The magnetic flux, Φ, through a surface is given by the dot product of the magnetic field, B, and the area vector through which the magnetic field lines are passing. Since the magnetic field is perpendicular to the plane of the ring, the magnetic flux can be expressed as: Φ = BAcosθ where A is the area of the ring, and θ is the angle between the magnetic field and the area vector. Since the ring is rotating about its diameter, we can write θ as a function of time: θ(t) = w*t where w is the angular velocity and t is the time.
03

Compute the rate of change of magnetic flux

Now, we need to find the rate of change of magnetic flux with respect to time. To do this, we take the derivative of Φ(t) with respect to t: \(\frac{dΦ}{dt} = \frac{d}{dt}( BAcos(wt))\) We can factor out B and A as they are constants: \(\frac{dΦ}{dt} = BA\frac{d}{dt}(cos(wt))\) Now, differentiate cos(wt) with respect to t: \(\frac{dΦ}{dt} = BA(-w \sin(wt))\) So, \(\frac{dΦ}{dt} = -BAw \sin(wt)\)
04

Calculate the induced emf

Now, we can use Faraday's law to calculate the induced emf: \(emf = -N \frac{dΦ}{dt}\) We have N = 20 turns, and we found the rate of change of magnetic flux in Step 3. Substituting the values, we get: \(emf = 20(-(-BAw \sin(wt)))\) Finally, we can simplify the expression: \(emf = 20 BAw \sin(wt)\) Recall that A = πr², where r is the radius of the ring. Plugging this into the equation, we get: \(emf = 20 B \pi r^2 w \sin(wt)\) So, the correct answer is (a) \(20 B \pi r^{2} \sin(wt)\).

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Most popular questions from this chapter

An inductor of inductance \(\mathrm{L}\) and resistor of resistance \(\mathrm{R}\) are joined in series and connected by a source of frequency 0 power dissipated in the circuit is, (a) $\left[\left\\{\mathrm{R}^{2}+\mathrm{c}^{2} \mathrm{~L}^{2}\right\\} / \mathrm{V}\right]$ (b) $\left[\left\\{\mathrm{V}^{2} \mathrm{R}\right\\} /\left\\{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}\right\\}\right]$ (c) $\left[\mathrm{V} /\left\\{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}\right\\}\right]$ (d) $\left[\sqrt{ \left.\left\\{R^{2}+\omega^{2} L^{2}\right\\} / V^{2}\right]}\right.$

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