(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.

https://tycho-s.phys.washington.edu/cgi/courses/shell/common/showme.pl?courses/phys122/autumn08/homework/09/rectangular_loop_MFR/p8.gif [Broken]

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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Magnetic field on current loop, involves rotational energy

**Physics Forums | Science Articles, Homework Help, Discussion**