Magnetic field on current loop, involves rotational energyby cdingdong Tags: current, kinetic energy, magnetic field, potential energy 

#1
Dec1608, 12:11 AM

P: 3

1. The problem statement, all variables and given/known data
A rectangular loop of sides a = 0.3 cm and b = 0.8 cm pivots without friction about a fixed axis (zaxis) that coincides with its left end (see figure). The net current in the loop is I = 3.2 A. A spatially uniform magnetic field with B = 0.005 T points in the +ydirection. The loop initially makes an angle q = 35° with respect to the xz plane. The moment of inertia of the loop about its axis of rotation (left end) is J = 2.9 × 106 kg m2. If the loop is released from rest at q = 35°, calculate its angular velocity [tex]\omega[/tex] at q = 0°. 2. Relevant equations this is kinetic energy: KE = 1/2I[tex]\omega^{2}[/tex] but moment of inertia is specified to be J, so KE = 1/2J[tex]\omega^{2}[/tex] this is potential energy for a loop with current and magnetic field acting on it: PE = [tex]\mu[/tex]Bcos[tex]\theta[/tex] moment vector [tex]\mu[/tex] = NIA where N = number of loops, I = current, A = area, B = magnetic field so, PE = NIABcos[tex]\theta[/tex] 3. The attempt at a solution I recognize that the angular kinetic energy equation can be used to find the angular velocity. KE = 1/2J[tex]\omega^{2}[/tex]. Final kinetic energy = initial potential energy. Initial potential energy = NIABcos[tex]\theta[/tex]. so, we could say 1/2J[tex]\omega^{2}[/tex] = NIABcos[tex]\theta[/tex] the right side is potential energy. that equalsso, now we have 1/2J[tex]\omega^{2}[/tex] = 3.2766e7 J = 2.9 × 106 kilogram meters squared. we'll call this 2.9e6 1/2(2.9e6)[tex]\omega^{2}[/tex] = 3.2766e7 [tex]\omega[/tex] = [tex]\sqrt{(2*3.2766e7)/2.9e6}[/tex] = 4.753655581 * 10^1 BUT, the answer happens to be what I did except that they did not take the square root at the end. so, they got 2.259724138 * 10^1. did i do something wrong? what i did makes sense in my mind. is their answer wrong? why did they not not take a square root at the end? is kinetic energy not = 1/2J[tex]\omega^{2}[/tex], but instead 1/2J[tex]\omega[/tex] without the square? i thank everybody in advance! 



#3
Dec1708, 02:32 PM

P: 3

wow, now that you bring it up, i realize my mistake. you're right! i forgot the final potential energy. so, the answer works, but there is something that bothers me.
K_{i} + U_{i} = K_{f} + U_{f} there is no initial kinetic energy because it came from rest, so that is zero U_{i} = K_{f} + U_{f} rearranging it, U_{i}  U_{f} = K_{f} NIABcos35  NIABcos0 = 1/2J[tex]\omega^{2}[/tex] NIAB(cos35  cos0) = 1/2J[tex]\omega^{2}[/tex] (1)(3.2)(2.5e5)(0.005)(cos35  cos0) = 1/2(2.9e5)[tex]\omega^{2}[/tex] 7.233918228 = 1/2(2.9e5)[tex]\omega^{2}[/tex] when i find the potential energy, initial PE  final PE, i get a negative number. so, when i set it equal to 1/2J[tex]\omega^{2}[/tex], i get a square root of a negative number, which is not possible. if i take the square root of the absolute value of that number, the answer is correct. is there something wrong with my signs? 



#4
Dec1708, 05:39 PM

Mentor
P: 40,889

Magnetic field on current loop, involves rotational energySo Ui  Uf = (IABcos35)  (IABcos0) = IAB(cos0  cos35). 



#5
Dec1808, 12:16 PM

P: 3

ahhhh, i see. well, that will solve my problem. thanks Doc Al!



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