Answer:
The emf induced in the ring increases by a factor of 2.
Explanation:
The formula of induced emf is given by :
[tex]\epsilon=-\dfrac{d\phi}{dt}\\\\\epsilon=-\dfrac{d(BA)}{dt}\\\\\epsilon=-\pi r^2\dfrac{dB}{dt}[/tex]
If the the radius of the ring is doubled, while the rate of increase of the field is cut in half.
New emf is given by :
[tex]\epsilon=-\pi r'^2\dfrac{dB'}{dt}\\\\\epsilon'=-\pi (2r)^2\dfrac{dB}{2dt}\\\\\epsilon'=-2\pi r^2\dfrac{dB}{dt}\\\\\epsilon'=-2\times \epsilon[/tex]
So, the emf induced in the ring increases by a factor of 2.
