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A metal ring 4.00 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 0.210 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?

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mavila18

Answer:

a) 0.21N/C

b) counterclockwise

Explanation:

a) to find the magnitude of the electric field you can use the following formula:

[tex]\int Eds=-\frac{\Delta \Phi_B}{\Delta t}=-\frac{\Delta AB}{\Delta t}[/tex]

A: area of the ring = pi*r^2

E: electric field

Ф_B: magnetic flux

In the line integral you can assume E as constant. Furthermore, you calculate the change in the magnetic flux by taking into account that the time interval is 1.12/0.21=5.33s. By replacing in the formula you obtain:

[tex]\frac{\Delta \Phi_B}{\Delta t}=\frac{A(B_f-B_i)}{5.33s}=\frac{\pi(0.04m)^2(1.12T)}{5.33}=1.056*10^{-3}W/s[/tex]

[tex]E\int ds=E(2\pi r)=1.056*10^{-3}W/s\\\\E=\frac{1.056*10^{-3}W/s}{\pi(0.04m)^2}=0.21\frac{N}{C}[/tex]

the magnitude if the induced electric field is 0.21N/C

b) By the Lenz's law you can conclude that the current has a direction in a counterclockwise

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