A coil of 43 turns in the shape of a rectangle with width 5 cm and length 10 cm is dropped from a position where magnetic field is 0 T to a position where the field is 0.5 T and is directed perpendicular to the plane of the coil. The displacement occurs in a time of 0.431 s. Calculate the resulting average emf induced in the coil

Explanation:

Given that,

Number of turns

N = 43 turns

Dimensions of rectangle

Length L = 10cm = 0.1m

Width W = 5cm = 0.05

Then, area of rectangle

A = L×W = 0.1 × 0.05 = 0.005m²

Initial magnetic field

Bi = 0T

Final magnetic field

Bf = 0.5T

The displacement time

∆t = 0.431s

Induce emf ε?

EMF is given as

ε = —NdΦ/dt

Where magnetic flux

Φ = BA

Note A is constant

Then,

ε = —NdΦ/dt = —Nd(BA)/dt

ε = —N•A•dB/dt

ε = —N•A•(∆B/∆t)

ε = —N•A•(Bf - Bi) / (∆t)

ε = —43 × 0.005 × (0.5 - 0) /0.431

ε = —43 × 0.005 × 0.5 /0.431

ε = —0.249 V

The average induced EMF is 0.249V

explanation:

higher potential

explanation:

because it is in a high speed so it must move into a of high potential