Question

A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant rate of 12.0cm/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop of magnitude 0.500T. You may want to review (Pages 957 - 964) . For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Generator iii: the slidewire generator.

Answer 1

Answer:

see explanation

Explanation:

Let me fill the rest of the question for you, it is a problem where you are supposed to find the electromotive force when a certain amount of time $t$ has passed, additionally, you have to find the direction of the induced current in the loop...

So you have a circular loop made out of iron with a circumference equal to 1.65 [m], (this is only so we can get the cross section area of the loop to fit it in an equation) and the decreasing rate of such is 0.12 [m/s].

Now a constant magnetic field which magnitude is 0.5 [T]  is oriented perpendicular to the ring, we have to define whether if it is entering the plane of the ring or going outside,

If the magnetic field is entering the plane, by the right hand rule we have that the currrent is going  clockwise, if the magnetic field is going outside it will go counter-clockwise

Now let's solve the problem

let's get the cross section area of the loop

we have the perimeter $P=2\pi r$ where r is the radius of the circle, we have P which is 1.65 [m], then $r = \frac{P}{2\pi }$

now the area A is $A = \pi r^2=\pi \frac{P^2}{4\pi^2}= \frac{P^2}{4\pi }$

Next we have that the EMF is the derivative of the flux with respect to t, and flux is $\Phi = \int \vec B. \vec{dA}$, so B and A are in the same direction so that is equal to

$\Phi = \int B*dA*cos(0) =B* \int dA = BA=B*\frac{P^2}{4\pi }$

next the EMF $\epsilon$ will be $|\epsilon| = |\frac{d}{dt }\Phi | = | \frac{B}{4\pi } \frac{d}{dt}(P^2)| = |\frac{BP}{2\pi } \frac{dP}{dt}|$

now the perimeter will be a function of t

$P(t) = P - \frac{dP}{dt}*t$

you just need to replace the values in order to get the answer

let's say that $t=5 [s]$

so we will have $P(t) = 1.65 - 0.12*5=1.05[m]$

next we replace in the emf equation:

$|\epsilon(t=5)| =|\frac{(0.5[T])*(1.05[m])}{2\pi }*0.12[m/s]| = 10 mV$