A circular loop of flexible iron wire has an initial circumference of

Monique Slaughter

Monique Slaughter

Answered question

2021-12-17

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.

Answer & Explanation

Donald Cheek

Donald Cheek

Beginner2021-12-18Added 41 answers

Part (a): The induced emf around a circular coil is given by
E=dϕdt
Where, ϕ is the magnetic flux passing through the surface area of the coil, which is given by the following equation
ϕ=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=dBAdt=BdAdtAdBdt
And since the magnetic field is constant through time, then we have
E=BdAdt0
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
2πr
Hence, the initial radius of the circular coil is thus
r=165.02π=262606cm
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
ddt2πr=2πdrdt=12cms
The negative sign indicates that the circumference is decreasing, dividing both sides by 2π, we get
drdt=1.90986cms
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=πr2
Hence differentiating the area with respect to time, we get
dAdt=2πrdrdt
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
drdt=1.90986
Hence, the rate of change of the surface of the coil is thus
dAdt==2πrx=1.90986
And, plugging in the equation of the radius, we get
dAdt=2π(1.90986t+26.2606)x1.90986
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
dAdtmidt9=2π(1.90986x9+26.2606)x1.90986108.862cm2s
Converting this into meter square per second by multiplying by 104, we get
dAdtmidt9=0.0108862m2s
Hence the emf after 9 seconds, is thus,

Neil Dismukes

Neil Dismukes

Beginner2021-12-19Added 37 answers

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.
dCdt=0.12ms
d(2πr)dt=012ms
drdt=0.019ms
The induced emf is given by :
ϵ=d(BA)dt
ϵ=2πBrdrdt
ϵ=2π×0.5×0.825×0019
ϵ=00492
(B) the direction of induced emf is clockwise as viewed looking along the direction of the magnetic fieldinside the loop.

nick1337

nick1337

Expert2021-12-27Added 777 answers

We are aware of what the loop's emf is ddt
here =BACosθ
Since the magnetic field is uniform here, we may claim that angle is zero
emf=d(BAcos(0))dt
emf=BdAdt
so area will be
dAdt=d(πr2)dt
dAdt=2πrdrdt
we know 2πr=c
r=c2π
r=1652π
3) emf=B2πrdrdr
and
Here, we discover the radius's rate of change, which is
drdt=122π=1.91θ102cm/s
therefore, after 9.0s have passed that coil's radius = 26.27 - 191 (9)
radius=9.08102cm
so now from equation 3 we find emf
emf=(0.500)x2π(9.08)/1.91
emf=0.005445V
and magnitude of emf = 0.005445 V
so
According to the right-hand rule, the direction of the current in this case is clockwise since the induced emf is 0.005445 V and the area is decreasing, which causes the flux to also be decreasing.

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