Calculate the induced electric field (in V/m) in a 40-turn coil with a diameter of 11 cm that is placed in a spatially uniform magnetic field of magnitude 0.35 T so that the face of the coil and the magnetic field are perpendicular. This magnetic field is reduced to zero in 0.10 seconds. Assume that the magnetic field is cylindrically symmetric with respect to the central axis of the coil. (Enter the magnitude.)

Explanation:

Given that,

Number of turn N = 40

Diameter of the coil d= 11cm = 0.11m

Then, radius = d/2 = 0.11/2 =0.055m

r = 0.055m

Then, the area is given as

A =πr²

A = π × 0.055²

A = 9.503 × 10^-3 m²

Magnetic Field B = 0.35T

Magnetic field reduce to zero in 0.1s, t = 0.1s

so we want to find induce electric field. To find the electric field,(E) we need to find the electric potential (V).

E.M.F is given as

ε = —N • dΦ/dt

Where magnetic flux is given as

Φ = BA

Then, ε = —N • dΦ/dt

ε = —N • dBA/dt

ε = —NBA/t

Then, its magnitude is

ε = NBA/t

Inserting the values of N, B, A and t

ε = 40×0.35×9.503×10^-3/0.1

ε = 1.33 V

Then, using the relationship between Electric field and electric potential

V = Ed

ε = E•d

E = ε/d

E = 1.33/0.11

E = 12.09 V/m

Thank you again :D