Answer :
Answer:
The magnitude of the average induced emf in the wire during this time is 9.533 V.
Explanation:
Given that,
Radius r= 0.63 m
Magnetic field B= 0.219 T
Time t= 0.0572 s
We need to calculate the average induce emf in the wire during this time
Using formula of induce emf
[tex]E=-\dfrac{d\phi}{dt}[/tex]
[tex]E=-B\dfrac{dA}{dt}[/tex]
[tex]E=-B\dfrac{A_{2}-A_{1}}{dt}[/tex]
[tex]E=B\dfrac{A_{1}-A_{2}}{dt}[/tex].....(I)
In reshaping of wire, circumstance must remain same.
We calculate the length when wire is in two loops
[tex]l=2\times 2\pi\times r_{1}[/tex]
[tex]l=2\times 2\pi\times 0.63[/tex]
[tex]l=7.916\ m[/tex]
The length when wire is in one loop
[tex]l=2\pi\times r_{2}[/tex]
[tex]7.916=2\times \pi\times r_{2}[/tex]
[tex]r_{2}=\dfrac{7.916}{2\times \pi}[/tex]
[tex]r_{2}=1.259\ m[/tex]
We need to calculate the initial area
[tex]A_{1}=N\times\pi\times r_{1}^2[/tex]
Put the value into the formula
[tex]A_{1}=2\times3.14\times(0.63)^2[/tex]
[tex]A_{1}=2.49\ m^2[/tex]
The final area is
[tex]A_{2}=N\times\pi\times r_{2}^2[/tex]
[tex]A_{2}=1\times\pi\times(1.259)^2[/tex]
[tex]A_{2}=4.98\ m^2[/tex]
Put the value of initial area and final area in the equation (I)
[tex]E=0.219\dfrac{2.49-4.98}{0.0572}[/tex]
[tex]E=-9.533\ V[/tex]
Negative sign shows the direction of induced emf.
Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.