Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

induced emf in a pendulum

1 vote
1,107 views

Q: a simple pendulum with Bob of mass m and conducting wire of length L swings under gravity through an angle $2\theta $. the Earth's magnetic field component in direction perpendicular to the plane of Swing is B, maximum potential difference induced across the pendulum would be ?
The figure and options for question are as follows :

I thought $E=d\phi_B/dt $ where $\phi_B $ is magnetic flux,
so $E= B× dA/dt $
Now I can't figure out how to find the area which is changing ?

asked Apr 8, 2017 in Physics Problems by physicsapproval (2,320 points)
edited May 4, 2017 by physicsapproval

1 Answer

1 vote
 
Best answer

There is no loop (closed circuit) in this problem, so you cannot use Faraday's Law. You must use the fact that the emf across a wire of length L moving at velocity v perpendicular to a magnetic field B is BLv.

The difficulty is that different segments of the conducting wire move at different speeds. So you must integrate from the pivot to the bob, adding the contribution from successive segments.

For all points on the wire, the maximum speed occurs when the wire passes through the vertical.


The amplitude of swing is $\theta$ so the maximum speed of the bob is given by
$v_0^2=2gh=2gL(1-\cos\theta)=4gL\sin^2\frac{\theta}{2}$
$v_0=2\sqrt{gL}\sin\frac{\theta}{2}$.

The emf induced in a wire of length $L$ moving with speed $v$ perpendicular to a field $B$ is $BLv$. Speed $v$ in the pendulum wire varies uniformly from 0 to $v_0$ so the average speed is $\frac12 v_0$. Because $V \propto v$ and $v$ varies uniformly we can use average speed instead of integrating.

The total emf induced in the pendulum wire is
$V=\frac12 BLv_0=BL\sqrt{gL}\sin\frac{\theta}{2}$.


Note: All of the options have different dimensions. So a quick way of getting the answer is to check which option has the dimensions of voltage - assuming that this option is correct.

answered Apr 8, 2017 by sammy gerbil (28,746 points)
edited Apr 10, 2017 by sammy gerbil
what I have done is $ mgx(1-cos\theta ) = 1/2 m v^2$ now how should I integrate for velocity ?
well, the dimentional analysis is short and quick  which I did in my test :)
The expression you have gives the velocity $v_L$ of the bob as it passes through the vertical. (x=L.) The velocity of the wire varies from 0 at the pivot to $v_L$ at the bob. The emf has to be integrated using $dV=Bv(x)dx$ where $x$ is distance from pivot and $v(x)=\frac{x}{L}v_L$.
how $v (x)= (x/L)v_L$ comes ?
Because the velocity of segments of wire increases linearly from pivot (x=0) to bob (x=L).
okay got that then that is because of $v_{max} = wA $ where $w $ is angular frequency and A be amplitude ?
Yes $v=\omega x$ and $v_{max}=v_L=\omega L$. Here $\omega=\frac{v_{max}}{L}$.
...