 # A circular loop of flexible iron wire has an initial circumference of Monique Slaughter 2021-12-17 Answered
A circular loop of flexible iron wire has an initial circumference of 165.0 cm, but its circumference is decreasing at a
constant rate of 12.0 cm/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 and with magnitude 0.500 T.
(a) Find the emf induced in the loop at the instant when 90 s have passed.
(b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
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Part (a): The induced emf around a circular coil is given by
$E=-\frac{d\varphi }{dt}$
Where, $\varphi$ is the magnetic flux passing through the surface area of the coil, which is given by the following equation
$\varphi =BA$
This means that the emf is induce across the coil, only if the coil experience a change in its value with respect to time.
Normally the magnetic field should be varying and the circular coil should have a constant surface area, but this is not the case in this problem, instead it is given that the magnetic field is constant, thus if the surface area of the coil loop were constant, this would mean that the induced emf force would be zero, differentiating the magnetic flux,we have
$E=-\frac{dBA}{dt}=-B\frac{dA}{dt}-A\frac{dB}{dt}$
And since the magnetic field is constant through time, then we have
$E=-B\frac{dA}{dt}-0$
Now it is given that the circumference is changing by a rate of 12 cm/s, and that the circumference of the coil is 165.0 cm, and since the circumference is given by
$2\pi r$
Hence, the initial radius of the circular coil is thus
$r=\frac{165.0}{2\pi }=\frac{26}{2606}cm$
And since it is given that the decreasing with respect to time with a rate of 12 cm/s, then we differentiate the circumference with respect to time, and hence we have
$\frac{d}{dt}2\pi r=2\pi \frac{dr}{dt}=-12c\frac{m}{s}$
The negative sign indicates that the circumference is decreasing, dividing both sides by $2\pi$, we get
$\frac{dr}{dt}=-1.90986c\frac{m}{s}$
Now from equation (1), we need to find the rate by which the area of the circular loop changes with time in order
to be able to find the induced emt, and from equation (2) we can integrate in order to find an equation
representing the value of the radius of the circular at any time instance, hence integrating we get
$r=-1.90986t+C$
And at time t = 0 , we know that the initial radius of the circular coil is 26.2606 cm, hence plugging in t = 0, we get
$26.2606=0+C$
Hence, the constant of integration is thus
$C=26.2606$
Hence, the equation representing the radius at any time instance is thus
$r=-190986t+26.2606$
And the area of a circular loop is given by
$A=\pi {r}^{2}$
Hence differentiating the area with respect to time, we get
$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$
And the rate at which the radius is decreasing is constant through the whole interval of time, thus we can plug in
the above equation the value of the rate of change of the radius wrt time, i.e. plugging in
$\frac{dr}{dt}=-1.90986$
Hence, the rate of change of the surface of the coil is thus
$\frac{dA}{dt}==2\pi rx=1.90986$
And, plugging in the equation of the radius, we get
$\frac{dA}{dt}=2\pi \left(-1.90986t+26.2606\right)x—1.90986$
In fact the above equation represents not the rate of change of the area, it rather expresses how much the area of
the coil loop have decreases at any instance of time.
And since the it is required to find how much induced emf are produced across the coil after 9 seconds, then we
need to find by how much the area would decrease in 9 seconds, hence substituting by t = 9, we get
$\frac{dA}{dt}mi{d}_{t-9}=2\pi \left(-1.90986x9+26.2606\right)x1.90986-108.862c\frac{{m}^{2}}{s}$
Converting this into meter square per second by multiplying by ${10}^{-4}$, we get
$\frac{dA}{dt}mi{d}_{t-9}=0.0108862\frac{{m}^{2}}{s}$
Hence the emf after 9 seconds, is thus,

###### Not exactly what you’re looking for? Neil Dismukes

Explanation:
Its given that,
Initial circumference of a circular loop, C = 165 cm = 1.65 m
Radius, $r=0.825m$
Time, $t=9s$
(A) The circumference is decreasing at a constant rate of 12 cm/s due to a tangential pull on the wire.
$\frac{dC}{dt}=-0.12\frac{m}{s}$
$\frac{d\left(2\pi r\right)}{dt}=-\frac{0}{12}\frac{m}{s}$
$\frac{dr}{dt}=0.019\frac{m}{s}$
The induced emf is given by :
$ϵ=-\frac{d\left(BA\right)}{dt}$
$ϵ=2\pi Br\frac{dr}{dt}$
$ϵ=2\pi ×0.5×0.825×\frac{0}{019}$
$ϵ=\frac{0}{0492}$
(B) the direction of induced emf is clockwise as viewed looking along the direction of the magnetic fieldinside the loop.

###### Not exactly what you’re looking for? nick1337

We know that the emfin the loop is $\frac{d\mathrm{\varnothing }}{dt}$
here $\mathrm{\varnothing }=BACos\theta$
so we say the that angle is zero and magnetic field is uniform here so that.
$emf=d\frac{\left(BA\mathrm{cos}\left(0\right)\right)}{dt}$
$emf=-B\frac{dA}{dt}$
so area will be
$\frac{dA}{dt}=\frac{d\left(\pi {r}^{2}\right)}{dt}$
$\frac{dA}{dt}=\frac{2\pi rdr}{dt}$
we know $2\pi r=c$
$r=\frac{c}{2\pi }$
$r=\frac{165}{2\pi }$
3) $emf=B\frac{2\pi rdr}{dr}$
and
here we find rate of change of radius that is
$\frac{dr}{dt}=\frac{12}{2\pi }=1.91\theta {10}^{2}cm/s$
so when 9.0s have passed that radius of coil = 26.27 - 191 (9)
$radius=9.08\ast {10}^{2}cm$
so now from equation 3 we find emf
$emf=-\left(0.500\right)x2\pi \left(9.08\right)/1.91$
$emf=-0.005445V$
and magnitude of emf = 0.005445 V
so
emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise.