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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.

Respuesta :

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]