How is EMF induced in a pendulum under the influence of Earth's magnetic field?

[nishanth R]

Homework Statement

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 the direction perpendicular to swing is B. The maximum potential difference induced across the pendulum is

Homework Equations

MAGNETIC FLUX=BAcos theta

The Attempt at a Solution

Area traversed by pendulum in 2 theta=theta * L^2
Magnetic flux=BA=B*Theta*L^2
EMF=d(B*Theta*L^2)/dt
=BL^2*d(theta)/dt
=BL^omega
=BL^2*(g/L)^1/2

But the answer is BL*sin (theta/2)*(gl)^1/2

 

[Ananya0107]

Mgl(1-cosθ) = 1/2 ml² ω²
Which gives ω= (2g(1-cosθ)/l)^½
Which gives E = Bωl² /2
E = Bl²sin(θ/2) (g/l)^½

 

[Ananya0107]

I think this is the maximum emf produced because when the pendulum rotates through an angle θ the area swept by it first decreases then increases again for the other half of the motion. Therefore the maximum emf would be produced when it sweeps through an angle θ.

 

[Coffee_]

You use $\dot{\theta}=w$ while $\dot{\theta}$ is something different here as pointed out earlier - we are not assuming uniform circular motion here remember.

 

[Ananya0107]

Hey

You use $\dot{\theta}=w$ while $\dot{\theta}$ is something different here as pointed out earlier - we are not assuming uniform circular motion here remember.

Is my method right?

 

[Coffee_]

Hey

Is my method right?

You are right about there being a maximum emf, the original commenter assumes a constant $\dot{\theta}$ which is actually a function of time for the harmonic oscillator, the maximal value of this derivative depends on the initial angle of release as the first commenter pointed out.

 

[nishanth R]

But if I replace the area to be transversed as theta, then is my approach correct?

Area traversed by pendulum in theta=(theta * L^2)/2
Magnetic flux=BA=B*(Theta*L^2)/2
EMF=d(B*Theta*L^2)/dt
=(BL^2)/2*d(theta)/dt
=(BL^2)/2omega
=BL^2/2*(g/L)^1/2

But the answer is BL*sin (theta/2)*(gl)^1/2

 

[nishanth R]

Coffee_ said: ↑
You use θ˙=w while θ˙ is something different here as pointed out earlier - we are not assuming uniform circular motion here rememb

I suppose omega is constant in pendulum motion. It is equal to (g/L)^1/2

 

[Coffee_]

Coffee_ said: ↑
You use θ˙=w while θ˙ is something different here as pointed out earlier - we are not assuming uniform circular motion here rememb

I suppose omega is constant in pendulum motion. It is equal to (g/L)^1/2

$w$ is indeed constant, but for a harmonic oscillator $\dot{\theta}(t)$ is not the same as $w$. Remember that $\theta(t)=Acos(wt+\phi)$ and so the derivative is still a function of time and not constant. You are correct up to the point of $EMF=\frac{BL^{2}}{2}\frac{d\theta}{dt}$ from this point on you make a mistake. What you have to do is realize that EMF is a function of time because $\frac{d\theta}{dt}$ is also a function of time. So what you have to do now is find the maximal value of $\frac{d\theta}{dt}$ - easiest way to do so is to use conservation of energy from the initial position.

EDIT : To do so consider the relationship $r\dot{\theta}=v$ , and when v is maximal, $\dot{\theta}$ is maximal as well.