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Homework Help: B-field generated by inducting loop of shrinking size (NEED SOMEONE TO CHECK ANSWERS)

  1. Mar 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A little stuck on the second question it basically boils down to not quite understanding the question. Was hoping someone here could push me in the right direction. The Question is two part, already completed the first half but wouldn't mind if someone looked over it while they are at it.

    The above conductor between points a and c on figure 4 lay in the paper's plane and create a circular loop with a shrinking radius r(t). The points a and c are moving away from each other such that the radius shrinks with a constant speed of r(t) = r0 - alpha*t for the time 0 < t < r0/alpha. There is not contact between the wires where they cross at the loop. An outer magnetic field, B > 0 pointing perpendicular into the paper's plane causes the loop to induce an electromotive force E as it shrinks. The magnetic field is only present over the conductor that goes from a to c. The resistance R connects the points a and c via the lower conductor.

    Additional descriptive text for q

    It can be assumed, that the B-field does not induce further electromotive forces in the circuit and the circuits self-induction can be ignored.

    2. Relevant equations
    Question 1
    a.) Determine the size of the induced electromotive force E in the loop for 0 < t < r0/alpha

    b.) Determine the size of the total magnetic field Btot in the center of the loop for 0 < t < r0/alpha
    3. The attempt at a solution

    The size of the electromotive force can be found from Faraday's law of induction:

    \mathcal{E} = -\frac{d\phi B}{dt} = -\frac{dBA(t)}{dt}

    Since the outer magnetic field B is independent of the time yet the area of the loop is, we can rewrite Faraday's law as:

    \begin{equation} \label{EQ1: Faraday's modified}
    \mathcal{E} = -B\frac{dA(t)}{dt}

    The area of the loop is given by the area of a circle:

    A(t) = \pi r(t)^2 = \pi(r_0 - \alpha t)^2

    The derivate of the area with respect to $t$ is then:

    \frac{dA(t)}{dt} = -2\pi\alpha(r_0 - \alpha t)

    Inserting this into eq. \ref{EQ1: Faraday's modified} yields:

    \begin{equation} \label{ANS1: 1}
    \mathcal{E} = 2B\pi\alpha(r_0 - \alpha t) \qquad for \qquad 0 < t < \frac{r_0}{\alpha}

    As can be seen from eq. \ref{ANS1: 1} as the loop shrinks the induced force also shrinks.


    Not entirely sure what they mean by the total B-field for the loop's center. Is it not simply B > 0 as we were told earlier, but if that is the case then it would only be a matter of isolating for B from the solution to part a. That just doesn't seem right. I looked at Lenz's law a bit but it only seems to be a slight change to Faraday's Law stating that the induced emf always tends to cancel out the change that caused it, which isn't much help. Have I misunderstood the question? If someone can point me in the right direction that'd be great.

    Fixed latex error. Also is there a way to import latex completly instead of just equations, it won't let me import $equation$ equations.

    EDIT 2:
    I think I found a solution. I hope this is correct.

    From the Biot-Savart law we know for a loop that the B-field generated is:

    \begin{equation} \label{EQ2: Biot-Savar law}
    B = \frac{\mu_0 I}{2r(t)}

    From Ampere's law we know that the current can be found as:

    I = VR

    We found an expression for V in part b.), namely eq. \ref{ANS1: 1}. Using this equation we can find another expression for I, remembering we have the resistance R in the lower conductor:

    I = \frac{2B\pi\alpha(r_0 - \alpha t)}{R}
    Inserting this into eq. \ref{EQ2: Biot-Savar law} and adding the external B-Field we can find the total B-field:

    B_{tot} = B_{coil} + B_{external}

    B_{tot} = \frac{\mu_0 2B\pi\alpha(r_0 - \alpha t)}{2R(r_0 - \alpha t)} + B = \frac{\mu_0 B\pi\alpha}{R} + B

    B_{tot} = B(1 + \frac{\mu_0\pi\alpha}{R}) \qquad for \qquad 0 < t < \frac{r_0}{\alpha}

    From the above equation we can see that Btot is constant regardless of the size of the loop. This of course only applies within the time limit such that there actually is a loop.
    Last edited: Mar 2, 2016
  2. jcsd
  3. Mar 2, 2016 #2
    Found the solutions and an answer sheet from a classmate. The answers I wrote for both is correct. Leaving this up so others with a similar problem can find it.
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