Answer:
Induced emf is 0.324 V
Explanation:
We have,
Number of turns, n = 61.6
Radius of circular coil, r = 4.44 cm
Resistance of coil, R = 2.34 Ω
The magnitude of the magnetic field varies in time according to the expression :
[tex]B=a_1t+a_2t^2[/tex]
[tex]a_1=0.0411\\\\a_2=0.044[/tex]
The magnitude of the induced emf in the coil is given by :
[tex]\epsilon=\dfrac{d\phi}{dt}\\\\\epsilon=\dfrac{d(NBA)}{dt}\\\\\epsilon=NA\dfrac{dB}{dt}[/tex]
[tex]\epsilon=\pi r^2\dfrac{dB}{dt}[/tex]
At t = 9.21 s,
[tex]\dfrac{dB}{dt}=(0.0411+2\times 0.044 \times 9.21)\\\\\dfrac{dB}{dt}=0.851\ T/s[/tex]
[tex]\epsilon=61.6\times \pi \times (4.44 \times 10^{-2})^2\times 0.851[/tex]
[tex]\epsilon=5.27\times 10^{-3}\ V[/tex]
So, the magnitude of the induced emf in the coil is 0.324 V