What is the maximum torque on a square shaped wire with given parameters?

[davidbenari]

Homework Statement

Find the maximum torque on a square shaped wire provided N=120; l=4cm;R=10$\Omega$; B=0.2 T ; $\omega=200 revs/sec$

Homework Equations

$\tau=IaBsin(\theta)$
$\epsilon=\frac{d}{dt}\Phi$

The Attempt at a Solution

$i=\frac{\epsilon}{R}=-\frac{\frac{d}{dt} (BAcos\omega t)}{R}$
$i_{max}=\frac{NBA\omega}{R}$
$\tau=il^2 B sin(\omega t)$
$\tau_{max}=il^2B = 2.6x10^-5 Nm$

This solution depends on the fact that my area vector is initially parallel to the magnetic field.

 

[andrevdh]

Shouldn't it be since
the torque on the coil is τ = BANi and i = BANω/R we have that
∴ τ = (BAN)²ω/R
= 29,5x10^-3 Nm
?

 

[davidbenari]

I don't see why the term $\omega/R$ is in there (the equation for current). Could you please explain? thanks

 

[haruspex]

$i=\frac{\epsilon}{R}=-\frac{\frac{d}{dt} (BAcos\omega t)}{R}$
$i_{max}=\frac{NBA\omega}{R}$
$\tau=il^2 B sin(\omega t)$
$\tau_{max}=il^2B = 2.6x10^-5 Nm$

This is not valid. It is not the case that $\tau(t)=i_{max}l^2 B sin(\omega t)$. You cannot take the max value until you have the general equation for $\tau(t)$.

 

[andrevdh]

But you also have it ? ... i = emf/R = NBAω {sin(θ)} /R
That is the emf is the time derivative of the magnetic flux is NBAω sin(θ)
I think what haruspex is saying is that there is another time varying term
for the torque. So you have a sin(θ)cos(θ) term in the general equation for the torque?