11. P=V2/R;
EMF=dφ/dt
=d/dt(BAcosωt)
=(BÎ R2sin wt)/2w
P=V2/R
=(B2Î 2R4sin2wt)/4w2
Ans.B2Î 2R4/8w2
Q1. In a uniform magnetic field of induction B, a wire in the form of a semicircle of radius r rotates about the diameter of the circle with an angular frequency ω. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R then calculate the mean power generated per period of rotation.
Q2. Faraday's laws are a consequence of conservation of
a) Energy field (What is it?) b) Energy and magnetic field c) Charge d) Magnetic Field
Q3. A solenoid is 1.5m long and its inner diameter is 4.0cm. It has three layers of windings of 1000 turns each and carries a current of 2.0 amperes. Calculate the nearby value of magnetic flux for a cross section of the solenoid. Answer - 6.31 x 10-6 weber
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5 Answers
1
30@Ajoy - I'm afraid thats not the answer.
Anyway, I did the sum after a lot of tries.
Initially, let the angle between the area vector and magnetic field be 0.
Therefore at any time t,
\theta =\omega t
\phi =\int \vec{B}.\vec{dS}
\phi = BAcos\omega t = \frac{B\pi r^{2}}{2}cos\omega t
\xi = -\frac{d}{dt}(\phi )=\frac{B\pi r^{2}\omega}{2}sin\omega t
Now, total work done per period (T) of rotation,
W=\int_{0}^{T}{\frac{\xi ^{2}}{R}dt}
\begin{align*} W &=\int_{0}^{T}{\frac{B^{2}\pi^{2} r^{4}\omega ^{2}}{4R}\sin ^{2}\omega tdt} \\ &= \frac{B^{2}\pi ^{2}r^{4}\omega^{2}}{8R}\int_{0}^{T}{2\sin^{2}\omega tdt}\\ &= \frac{B^{2}\pi ^{2}r^{4}\omega^{2}}{8R}\int_{0}^{T}{\left( 1-\cos2\omega t\right)dt}\\ &= \left( \frac{B^{2}\pi ^{2}r^{4}\omega^{2}}{8R}\right)T \end{align*}
\text{Mean Power (P)}= \frac{W}{T}=\frac{\left( B \omega\pi r^{2}\right)^{2}}{8R}
1Oops......I wrote ω in the denominator in the differentiation step......otherwise its the same method....
