# magnitude of torque on a current loop

by ovoleg
Tags: current, loop, magnitude, torque
 P: 94 I've given this one a good effort and I cannot seem to solve it, been trying for a couple days on my own now...Anyone able to lend a hand? I was using t=uxB(u cross B) to find the torque and then take the negative of it to find the torque that is to hold it? vector u = I(vector A), and for this I am getting vector(A)=.0692i+.08j+.0346k vector(B)=.58i+0j+0k When I do the matrices for the cross(x) I get 0i+.321088j+.7424k The only part of this that is right is the i component which is zero... I'll be in your debt forever if you could help with a) or c) Thanks!!!! ------------------ The rectangular loop in Fig is pivoted about the y-axis and carries a current of 16.0 A in the direction indicated.((It's 16A, don't mind the picture saying 15A, this is the diagram below, click link)) a) If the loop is in a uniform magnetic field with magnitude 0.580 T in the +x-direction, find the magnitude and direction of the torque required to hold the loop in the position shown. = b) Repeat part (a) for the case in which the field is in the z-direction. = c) For each of the above magnetic fields, what torque would be required if the loop were pivoted about an axis through its center, parallel to the y-axis? a = b =
 P: 227 $$\tau = NIAB sin \Theta$$ N = number of loops I = current A = area of loop B = magnetic field theta = angle between loop and field
P: 94
 Quote by andrewchang $$\tau = NIAB sin \Theta$$ N = number of loops I = current A = area of loop B = magnetic field theta = angle between loop and field
I love you! why doesn't my book have this?? :*(.

Thanks alot!

P: 19

## magnitude of torque on a current loop

well it is sinQ if we take the angle of rotatation in the geomtrical center of the loop......

just for information..heh

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