Answer:
09m
Explanation:
yes the current flow in clockwise direction
Explanation:
Given that,
Diameter of the metal ring, d = 4.2 cm
Radius, r = 2.1 cm
Initial magnetic field, B = 1.12 T
The magnetic field is decreasing at the rate of, [tex]\dfrac{dB}{dt}=0.24\ T/s[/tex]
Due to change in magnetic field, an emf is induced in it. And hence, electric field is induced. It is given by :
[tex]\int\limits {E.dl}=\dfrac{d}{dt}(\int\limits{B.dA)}[/tex]
[tex]E.(2\pi r)=\dfrac{dB}{dt}(\pi r^2)[/tex]
[tex]E=\dfrac{dB}{dt}\times \dfrac{r}{2}[/tex]
[tex]E=0.24\times \dfrac{2.1\times 10^{-2}}{2}[/tex]
[tex]E=0.00252\ N/C[/tex]
So, the magnitude of the electric field induced in the ring has a magnitude of 0.00252 N/C.
The direction of electric field will be counter clock wise direction as viewed by someone on the south pole of the magnet.
