Calculating Induced Current in an Expanding Loop

[Andy13]

Homework Statement

Imagine a pliable round metal loop that can expand or contract. In a region with a constant magnetic field B0 that is oriented perpendicular to the plane of the loop, suppose that the loop expands, with its radius growing with time as r = r0(1+at2). As the loop expands and grows thinner, its resistance per unit length changes according to R = R0(1+bt2). Find an expression for the current induced in the loop as a function of time. To check your answer, suppose that B0 = 3.30 mT, r0 = 16.5 cm, R0 = 7.55Ohm/m, a = 0.245 x 10-4 s2, and b = 0.590 x 10-2 s2. What is the value of the induced current at t = 27.5 s? (Note: Give the direction of the current where when viewed from above a positive current will move counterclockwise.)

Homework Equations

V = IR
EMF = d(flux)/dt
flux = Sb*dA (integral of dot product of B and dA, or |B||dA|cos(90) in this case --> BdA)

The Attempt at a Solution

First, find EMF induced and then use V = IR to solve for I. This is the equation I came up with:

I(t) = B∏(ro(1+at^2))^2/(Ro(1+bt^2) = 7E-6 A

This answer wasn't correct in my online homework program. The negative value was also incorrect. Then, I reread the question and saw the part about the equation for resistance being per unit length, so I got this equation (multiply above by circumference):

I(t) = 2B∏^2(ro(1+at^2))^3/(Ro(1+bt^2) = 7E-6 A

It didn't change my answer. Which equation is right (if either)?

 

[phyzguy]

Neither one is correct. What is the induced EMF in the loop?

 

[Andy13]

Neither one is correct. What is the induced EMF in the loop?

Should be -d(flux)/dt = integral (B*dA)
the derivative of an integral of an equation with respect to t is just B*dA

So my mistake was to forget to differentiate at all. Whoops.

A = ∏r^2 but r= ro(1+at^2)
so dA = derivative of ∏(ro(1+at^2)^2 = 4∏*a*t*ro(1+at^2)

EMF = B*4∏*a*t*ro(1+at^2)

Does my derivative look ok? And, was I right in thinking that in order to get the resistance in the equation V = IR you have to multiply by the circumference because the equation gives resistance/unit length?

 

[phyzguy]

OK, you're on the right track now. Your derivative looks OK, except it should be r0^2. And yes, you are right that since the resistance is pre unit length, you need to multiply by the circumference.

 

[Andy13]

OK, you're on the right track now. Your derivative looks OK, except it should be r0^2. And yes, you are right that since the resistance is pre unit length, you need to multiply by the circumference.

Thanks, got it!

 

[Steely Dan]

Should be -d(flux)/dt = integral (B*dA)
the derivative of an integral of an equation with respect to t is just B*dA

The right side of this equation is the flux:

\Phi = \int \vec{B} \cdot d\vec{A}

So, after evaluating \Phi(t), you would then take its time derivative to get the emf. But to get this flux, you only need to multiply the magnetic field by the area, since

\Phi = \int \vec{B} \cdot d\vec{A} = BA

if the magnetic field is perpendicular to the area and is uniform throughout the area. Then you do

\text{emf} = -\frac{d\Phi}{dt}.

In particular, none of this needs to involve taking the derivative of the area.

 

[phyzguy]

\Phi = \int \vec{B} \cdot d\vec{A} = BA

if the magnetic field is perpendicular to the area and is uniform throughout the area. Then you do

\text{emf} = -\frac{d\Phi}{dt}.

In particular, none of this needs to involve taking the derivative of the area.

Um...what? If B is constant, and phi = BA, then isn't d(phi)/dt = B * dA/dt? Isn't this taking the derivative of the area? What are you trying to say?

 

[Steely Dan]

Um...what? If B is constant, and phi = BA, then isn't d(phi)/dt = B * dA/dt? Isn't this taking the derivative of the area? What are you trying to say?

I'm correcting the OP's terminology. It looks like he has the correct expression for the magnitude of the emf but seems confused on the proper derivative and naming conventions.