Question

30. A long solenoid with n=10 turns per centimeter has a cross-sectional area of 5.0cm2 and carries a current of 0.25 A. A coil with five turns encircles the solenoid. When the current through the solenoid is turned off, it decreases to zero in 0.050 s. What is the average emf induced in the coil?

Answer 1

Induced emf = ε = 1.5 × 10^-5 V

Step-by-step explanation:

Relative permeability = µ = 4π × 10^-7 N/A^2

Number of turns per meter = n = (10/cm)(100cm/m) = 1000/m

Number of turns = N = 5

Area = A = 5 × 10^-4 m^2

Current = I = 0.25 A

Time = t = 0.050 s

We know that,

Induced emf = ε = µ NnAI/t = (4π × 10^-7 )(5)(1000)(0.25)( 5 × 10^-4 )/0.05

Induced emf = ε = 1.5 × 10^-5 V

Answer 2

average emf induced in the coil is 1.57 x
`10^(-5)` V

Step-by-step explanation:

Given data

n = 10 turns per centimeter = 1000 m^-1

N = 5

cross-sectional area A = 5.0 cm2 = 5.0 x 10^-4 m²

current = 0.25 A

to find out

average emf induced in the coil

solution

we will find emf by the given formula

emf = - (μ×N×n×A×ΔI ) / t ..................1

here

μ = 4π x 10^-7 and

N = 10

n = 1000

A = 5.0 x 10^-4

ΔI = 0.25

t = 0.050

put all value in equation 1

emf = - (μ×N×n×A×ΔI ) / t

emf = - (4π x 10^-7 ×5 ×1000× 5.0 x 10^-4 × 0.25 ) / 0.050

emf = 1.57 x
`10^(-5)` V

average emf induced in the coil is 1.57 x
`10^(-5)` V