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

`emf=-(1)/(2)kta^5`

Step-by-step 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 =(1)/(4)kt^2a^5`

We know that emf given as

`emf=-(d\phi )/(dt)`

`\phi =(1)/(4)kt^2a^5`

So

`emf=-(1)/(2)kta^5`