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 thi

Question

18 days ago

12

s 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.

1 answer:

Softa [2K] · Answer 1 · 2026-03-08T06:10:31+03:00

Softa [2K]18 days ago

5 0

Answer:

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

Explanation:

We have

B(y, t) = k y ³t²

To determine the total flux through the loop, we need to integrate over it

$\phi =\int B.dS$

Since the loop is square,

$\phi =\int B.dS$

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 the emf can be expressed as

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

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

Thus

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