Explanation:
Given that,
Dimensions of the rectangle is 0.52 m × 0.64 m
Magnetic field, B₁ = 2 T
Final magnetic field, B₂ = 0
Time taken, t = 0.49 s
The loop is inclined at an angle of 82° with respect to the normal to the plane of the loop.
(a) The average induced emf is given by :
[tex]\epsilon=-\dfrac{d\phi}{dt}[/tex]
[tex]\epsilon=-\dfrac{d(BA\ cos\theta)}{dt}[/tex]
[tex]\epsilon=-Acos\theta \dfrac{B_2-B_1}{t}[/tex]
[tex]\epsilon=-0.52\times 0.64\times cos(82)\times \dfrac{0-2}{0.49}[/tex]
[tex]\epsilon=0.189\ V[/tex]
(b) [tex]\epsilon=-\dfrac{d\phi}{dt}[/tex]
[tex]\epsilon=-\dfrac{d(BA\ cos\theta)}{dt}[/tex]
[tex]\epsilon=-B\ cos\theta \dfrac{dA}{dt}[/tex]
[tex]\dfrac{dA}{dt}=\dfrac{\epsilon}{-B\ cos\theta}[/tex]
[tex]\dfrac{dA}{dt}=\dfrac{0.189}{2\times cos(82)}[/tex]
[tex]\dfrac{dA}{dt}=0.679\ m^2/s[/tex]
Hence, this is the required solution.
