Complete question is;
A circular loop of flexible iron wire has an initial circumference of 161 cm , but its circumference is decreasing at a constant rate of 11.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.800 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.
Find the emf induced in the loop, at the instant when 9.0s has passed.
Answer:
Magnitude of EMF induced = 0.008686 V
Explanation:
We are given;
Initial circumference; C = 161 cm
rate of decrease of circumference; dC/dt = 11 cm/s
magnetic field; B = 0.8 T
time; t = 9 sec
To find emf in loop, we'll use the formula;
EMF = -d(BAcosθ)/dt
Now, we are told magnetic field is uniform, thus angle is zero.
Thus;
EMF = -d(BAcos0)/dt
cos 0 = 1
Thus;
EMF = -B(dA/dt)
so area will be
Area of the loop is (πr²)
Thus;
dA/dt = d(πr²)/dt
Differentiating with respect to r, we have;
dA/dt = 2πr(dr/dt)
Formula for circumference is;
C = 2πr
Making r the subject, we have;
r = C/2π
Thus;
R = 161/2π
r = 25.6239 cm
c is circumference so from equation 2
From EMF = -B(dA/dt) , let's replace dA/dt with 2πr(dr/dt) since dA/dt = 2πr(dr/dt)
Thus;
EMF = -2πrB(dr/dt)
From C = 2πr, we can derive the rate of increase in radius.
Thus;
dC/dt = 2π(dr/dt)
Plugging in the relevant values, we have;
dr/dt = 11/2π = 1.75 cm/s
At the instance when 9 s has passed, radius of the could is now;
r1 = r - 9(dr/dt)
r1 = 25.6239 - 9(1.75)
r1 = 9.8739 cm
We will need to plug in 0.8 for B, 9.8739 cm for r, 1.75 cm/s for dr/dt into;
EMF = -2πrB(dr/dt)
But, we have to convert cm to m.
Thus: 9.8739 cm = 9.8739 × 10^(-2) m
1.75 cm = 0.0175 m
Thus;
EMF = -2π × 9.8739 × 10^(-2) × 0.8 × 0.0175
EMF = -0.008686 V
For magnitude of emf, we will take the absolute value.
Thus;
Magnitude of EMF = 0.008686 V
