Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

charge flown through a loop due to induced emf

3 votes
207 views

a square loop of side=a has its center at ( 3/2 )a distance away from a infinite wire carrying current . If the loop and the wire are in the same plane initially and then if the loop rotates by 180 degrees then what is the total amount of charge which passes through any point ofvthe loop during rotation? The loop rotates about an axis parallel to the long wire.

what I did is that $q=\int I.dt=\frac {-d\phi}{r.dt}.dt $ and dt cancels so it should be just $\triangle\phi$ which would be final - initial flux giving -2B.A but answer doesn't come by this
where I am thinking wrong

asked Nov 15, 2016 in Physics Problems by physicsapproval (2,320 points)
retagged Nov 15, 2016 by physicsapproval
The formula for charge flown is $\triangle \phi /R$
B is not constant, it changes with r = distance from wire.  Perhaps you must integrate over the area of the loop, assuming two sides are parallel to the current-carrying wire.
but then how to make equation for that as the loop is rotating
Your answer does not take speed of rotation into account. It compares initial with final flux linkages after the loop has been rotated. Instantaneous current depends on rate of change of flux linkage, but total charge depends on the change in flux linkage.

1 Answer

3 votes
 
Best answer

Suppose the wire loop has resistance R. The emf induced in the wire loop by a change in magnetic flux linkage $\Phi$ is
$E= -\frac{d\Phi}{dt}$.
The current in the loop is
$i=\frac{E}{R}=-\frac{d\Phi}{Rdt}$.
Charge flowing is the integral of current wrt time :
$q=\int \frac{E}{R}dt=-\int \frac{1}{R}d\Phi=\frac{2\Phi_0}{R}$
because $\Phi$ has changed direction.

By definition $\Phi=\int BdA$. Here $dA=drdz$ where $a<r<2a$ is perpendicular to the wire and $0<z<a$ is parallel to it. The magnetic field around the wire is
$B=\frac{\mu_0 I}{2\pi r}$.
Therefore
$\Phi_0=\frac{\mu_0 I}{2\pi} \int_0^a dz \int_a^{2a} \frac{1}{r} dr=a\frac{\mu_0 I}{2\pi}ln(2)$
hence
$q=\frac{a\mu_0 I}{\pi R}ln(2)$.

answered Nov 18, 2016 by sammy gerbil (28,448 points)
selected Nov 20, 2016 by physicsapproval
ok I understood where I went weong,thanks
...