Question

The component of the external magnetic field along the central axis of a 18 18 turn circular coil of radius 27.0 27.0 cm decreases from 1.90 1.90 T to 0.500 0.500 T in 2.90 2.90 s. If the resistance of the coil is R = 4.50 R=4.50 Ω, what is the magnitude of the induced current in the coil?

Answer 1

Answer:

Induced current I = 0.44 A

Explanation:

given data

no of turn N = 18

radius  = 27.0 cm

resistance of the coil = 4.50 Ω

time = 2.90 s

magnetic field  = 1.90 T to 0.500 T

solution

we get Induced EMF that is

E = - dφ ÷ dt    ..............1

E = - N × A × (B2 - B1) ÷ Δt

E = N × A × (B1 - B2) ÷ Δt

area of cross section of the coil A = π × r²

area of cross section = 3.14 × (0.27)²

area of cross section = 0.228 m²

Initial magnetic field B1   = 1.9 T

and  

Final magnetic field B2   = 0.5 T

time      Δt = 3.7 s

put here value and we get

E  =  $\frac{18 \times 0.228 \times (1.9-0.5)}{2.90}$  

E  = 1.98 V

and  

Induced current I is express as

Induced current I = E  ÷ R    ................2

Induced current I = $\frac{1.98}{4.5}$

Induced current I = 0.44 A