Answer:
[tex]emf=-\dfrac{1}{2}kta^5[/tex]
Explanation:
Given that
B(y, t) = k y ³t²
To find the total flux over the loop we have to integrate over the loop
[tex]\phi =\int B.dS[/tex]
Given that loop is square,so
[tex]\phi =\int B.dS[/tex]
B(y, t) = k y ³t²
[tex]\phi =kt^2\int_{0}^{a}dx\int_{0}^{a}y^3dy[/tex]
[tex]\phi =\dfrac{1}{4}kt^2a^5[/tex]
We know that emf given as
[tex]emf=-\dfrac{d\phi }{dt}[/tex]
[tex]\phi =\dfrac{1}{4}kt^2a^5[/tex]
So
[tex]emf=-\dfrac{1}{2}kta^5[/tex]
