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Induced emf in a shape being stretched from two vertices

2 votes

I have tried the part a of this problem nearly 6 times. In my fifth and 6th attempt I got the same answer and I was confident about my solution but I am getting the wrong answer nonetheless.

Here's my attempt:

Call the vertical diagonal p and the horizontal diagonal q.

Let the upper and lower vertices move towards the centre with velocity
$v$ just as the sidewards vertices move away with $u$.
From constraint relation, $2v\cos \alpha \cos\theta= u \implies v =
\dfrac{u}{ \sin 2\theta}$
Area of the figure at any instant is given by

$A= \dfrac 1 2 pq$

$\dfrac{dA}{dt}= \frac 12 \left(p

Now note the following relations:

$\dfrac{dq}{dt}= 2u$

$\dfrac{dp}{dt}= -2v$

$p = 2a \sin \theta$

$q = 2a \cos \theta$

$\theta = 30^\circ$
Substituting, we get: $\dfrac{dA}{dt} = -au$
$\implies \epsilon = -\dfrac{d\phi}{dt}= -B\dfrac{dA}{dt}= Bau $

But answer is : $2Bau$

Please let me know my error.

asked Jul 24, 2018 in Physics Problems by Reststack (422 points)
edited Jul 25, 2018 by Reststack

1 Answer

2 votes
Best answer

Your mistake is in the derivation of the relation $$v=\frac{u}{\sin2\theta}$$ which should be as follows : $$p=2a\sin\theta, q=2a\cos\theta$$ $$\dot p=2a\dot\theta \cos\theta, \dot q=-2a\dot\theta \sin\theta $$ $$\frac{\dot p}{\dot q}=-\cot\theta$$ $$\dot q=2u, \dot p=-\dot q \cot\theta=-2u\cot\theta$$ So you ought to have obtained $$v=u\cot\theta$$ When $\theta=30^{\circ}$ then $$\sin\theta=\frac12, \cos\theta=\frac{\sqrt3}{2}, \cot\theta=\frac{\cos\theta}{\sin\theta}=\sqrt3$$ so $$p=a, q=a\sqrt3, \dot p=-2u\sqrt3, \dot q=2u$$ and therefore $$\dot A=\frac12 [p\dot q+\dot p q]=\frac12 [(a)(2u)+(-2u\sqrt3)(a\sqrt3)]=au-3au=-2au$$

answered Jul 24, 2018 by sammy gerbil (28,178 points)
selected Jul 25, 2018 by Reststack