Question

A metal ring 4.5 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 counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?

Answer 1

Answer:

$2.81 \times 10^{-3}\ N/C$

Explanation:

B = Magnetic field = 1.12 T

Diameter = 4.5 cm

$\dfrac{dB}{dt}=0.25\ T/s$

Induced electric field is given by

$\int E d l=\frac{d}{d t}\left(\int B \cdot d A\right)$

$E(2 \pi r)=\frac{d B}{d t}\left(\pi r^{2}\right)$

$\Rightarrow E=\frac{d B}{d t}\left(\frac{r}{2}\right)$

$\Rightarrow E=\frac{d B}{d t}\left(\frac{r}{2}\right)$

$\Rightarrow E=(0.250)\left(\frac{4.5 \times 10^{-2}}{4}\right)$

$\Rightarrow E=2.81 \times 10^{-3}\ N/C$

The magnitude of the electric field induced in the ring is $2.81 \times 10^{-3}\ N/C$

If the magnetic field is viewed by someone on the south pole of magnet then the direction of current will be counterclockwise