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Magnetic field on current loop, involves rotational energy

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cdingdong
#1
Dec16-08, 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 (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 equals

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
so, now we have

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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Doc Al
#2
Dec17-08, 01:22 PM
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Quote Quote by cdingdong View Post
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]
What's the final potential energy, when θ = 0?
cdingdong
#3
Dec17-08, 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.

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[tex]\omega^{2}[/tex]

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

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

-7.233918228 = 1/2(2.9e-5)[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?

Doc Al
#4
Dec17-08, 05:39 PM
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Magnetic field on current loop, involves rotational energy

Quote Quote by cdingdong View Post
is there something wrong with my signs?
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).
cdingdong
#5
Dec18-08, 12:16 PM
P: 3
ahhhh, i see. well, that will solve my problem. thanks Doc Al!


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