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 Tbetween them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.200 T/s .Part AWhat is the magnitude of the electric field induced in the ring?E=____________v/mPart BIn which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?1- Counterclockwise2- Clockwise

Answer 1

Step-by-step explanation:

We have a metal ring of diameter

d = 4.2cm = 0.042m

r = d/2 = 0.021m

And it is place between the north pole and south pole of a large magnet with the plane of it's area perpendicular to the magnetic field.

Given that the magnetic field is

B = 1.12 T

The rate of decrease of magnetic field is 0.2T/s, since it is decrease then,

dB/dt = -0.2 T/s

The induce electric field is given as,

From faradays law

ε = ∫E•dl = -dΦ/dt

Magnetic flux is given as

Φ = BA

Φ = πr²×B = πr²B

Also, ∫E•dl = E×2πr = 2πrE

So,

∫E•dl = -dΦ/dt

2πrE = -d(πr²B) / dt

r is a constant, then

2πrE = -πr² dB/dt

Divide both side by πr

2E = -r dB/dt

E = -r dB/dt / 2

E = -0.021 × -0.2 / 2

E = 0.0021 V/m

The magnetic field point from north to south pole and it is decreasing and this means that the magnetic flux is also decreasing, so the induce magnetic field must point in the same direction of the original magnetic field, so the induce current circulate counter-clockwise as viewed from the south pole