Question

A metal ring 4.20 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.240 T/s . (A) What is the magnitude of the electric field induced in the ring? (B) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?

Answer 1

Answer:

09m

Explanation:

yes the current flow in clockwise direction

Answer 2

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, $\dfrac{dB}{dt}=0.24\ T/s$

Due to change in magnetic field, an emf is induced in it. And hence, electric field is induced. It is given by :

$\int\limits {E.dl}=\dfrac{d}{dt}(\int\limits{B.dA)}$

$E.(2\pi r)=\dfrac{dB}{dt}(\pi r^2)$

$E=\dfrac{dB}{dt}\times \dfrac{r}{2}$

$E=0.24\times \dfrac{2.1\times 10^{-2}}{2}$

$E=0.00252\ N/C$

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.