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

3 months 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 [3K] · Answer 1 · 2026-03-08T06:10:31+03:00

Softa [3K]3 months 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

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