Question

A metal ring 4.50 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 0.250 T/s. (a) What is the magnitude of the electric field induced in the ring? (b) In which direction (clockwise or counter clockwise) does the current flow as viewed by someone on the south pole of the magnet?

Short answer

(a) The magnitude of the electric field induced in the ring is$2.81\times {10}^{-3}\mathrm{V}/\mathrm{m}$

(b) The current flows in the counter clockwise direction, as viewed by someone from the south pole of the magnet.

Step-by-step solution

Magnitude of the electric field

Give,

A metal ring with a diameter of d = 4.50cm, magnetic field with magnitude B = 1.12T and the rate of the gradual decrease is$\frac{\mathrm{dB}}{\mathrm{dt}}=0.250\mathrm{T}$

According to the Faraday’s law we get:

$$\begin{gathered} \mathrm{\epsilon }=\oint \overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{dl}} \\ =-\frac{\mathrm{d\varphi B}}{\mathrm{dt}} \end{gathered}$$

Now, the magnetic flux is :

$$\begin{gathered} \varphi =AB \\ =\pi r^{2}B \end{gathered}$$

Now, integrating the LHS we get:

$\oint \overrightarrow{E}.\overrightarrow{dl}=\left(2\pi r\right)$

Therefore,

$$\begin{gathered} E2\pi r=\frac{d(B\pi r^2}{dt} \\ =\pi r^2\frac{dB}{dt} \\ =\frac{0.0225m}{2}0.250T/s \\ =2.81\times {10}^{-3}V/m \end{gathered}$$

Therefore, the electric field is$2.81\times {10}^{-3}V/m$

Direction of the magnetic field

The magnetic field points from north to south and it is decreasing meaning the magnetic flux is decreasing too. Hence the induced magnetic field must be pointed towards the same direction of the original magnetic field. Therefore, the current circulates counter clockwise as viewed from the south pole.