(adsbygoogle = window.adsbygoogle || []).push({}); 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 (z-axis) 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 +y-direction. The loop initially makes an angle q = 35° with respect to the x-z plane.

The moment of inertia of the loop about its axis of rotation (left end) is J = 2.9 × 10-6 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

N = 1 turn; I = 3.2 Amps; A = 0.003*0.008 = 2.5e-5 meters; B = 0.005 Teslas; [tex]\theta[/tex] = 35 degrees

(1)(3.2)(2.5e-5)(0.005)cos(35) = 3.276608177 * 10^-7 Joules. we'll call this 3.2766e-7

1/2J[tex]\omega^{2}[/tex] = 3.2766e-7

J = 2.9 × 10-6 kilogram meters squared. we'll call this 2.9e-6

1/2(2.9e-6)[tex]\omega^{2}[/tex] = 3.2766e-7

[tex]\omega[/tex] = [tex]\sqrt{(2*3.2766e-7)/2.9e-6}[/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!

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# Magnetic field on current loop, involves rotational energy

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