Question

A square loop of wire, with sides of length a, lies in the first quadrant of the xy plane, with one corner at the origin. In this region, there is a non uniform time-dependent magnetic field B(y, t) = ky3t2 ˆz (where k is a constant). Find the emf induced in the loop.

Answer 1

Answer:

$emf=-\dfrac{1}{2}kta^5$

Explanation:

Given that

B(y, t) = k y ³t²

To find the total flux over the loop we have to integrate over the loop

$\phi =\int B.dS$

Given that loop is square,so

$\phi =\int B.dS$

B(y, t) = k y ³t²

$\phi =kt^2\int_{0}^{a}dx\int_{0}^{a}y^3dy$

$\phi =\dfrac{1}{4}kt^2a^5$

We know that emf given as

$emf=-\dfrac{d\phi }{dt}$

$\phi =\dfrac{1}{4}kt^2a^5$

So

$emf=-\dfrac{1}{2}kta^5$