Maximum torque on a wire

  • #1
461
16

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.
 

Answers and Replies

  • #2
andrevdh
Homework Helper
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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
?
 
  • #3
461
16
I don't see why the term ##\omega/R## is in there (the equation for current). Could you please explain? thanks
 
  • #4
haruspex
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##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)##.
 
  • #5
andrevdh
Homework Helper
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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?
 
Last edited:

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