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

1. Dec 16, 2008

### cdingdong

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 $$\omega$$ at q = 0°.

2. Relevant equations

this is kinetic energy:
KE = 1/2I$$\omega^{2}$$ but moment of inertia is specified to be J, so

KE = 1/2J$$\omega^{2}$$

this is potential energy for a loop with current and magnetic field acting on it:
PE = $$\mu$$Bcos$$\theta$$
moment vector $$\mu$$ = NIA where N = number of loops, I = current, A = area, B = magnetic field

so, PE = NIABcos$$\theta$$

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$$\omega^{2}$$.
Final kinetic energy = initial potential energy. Initial potential energy = NIABcos$$\theta$$. so, we could say

1/2J$$\omega^{2}$$ = NIABcos$$\theta$$

the right side is potential energy. that equals

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

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

1/2J$$\omega^{2}$$ = 3.2766e-7

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

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

$$\omega$$ = $$\sqrt{(2*3.2766e-7)/2.9e-6}$$ = 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$$\omega^{2}$$, but instead 1/2J$$\omega$$ without the square?

Last edited by a moderator: May 3, 2017
2. Dec 17, 2008

### Staff: Mentor

What's the final potential energy, when θ = 0?

3. Dec 17, 2008

### cdingdong

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.

Ki + Ui = Kf + Uf

there is no initial kinetic energy because it came from rest, so that is zero

Ui = Kf + Uf

rearranging it,

Ui - Uf = Kf

NIABcos35 - NIABcos0 = 1/2J$$\omega^{2}$$

NIAB(cos35 - cos0) = 1/2J$$\omega^{2}$$

(1)(3.2)(2.5e-5)(0.005)(cos35 - cos0) = 1/2(2.9e-5)$$\omega^{2}$$

-7.233918228 = 1/2(2.9e-5)$$\omega^{2}$$

when i find the potential energy, initial PE - final PE, i get a negative number. so, when i set it equal to 1/2J$$\omega^{2}$$, 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. Dec 17, 2008

### Staff: Mentor

Yes, your signs are messed up. The correct expression for PE is PE = -IABcosθ (note the minus sign).

So Ui - Uf = (-IABcos35) - (-IABcos0) = IAB(cos0 - cos35).

5. Dec 18, 2008

### cdingdong

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