A solenoid of radius 1.50 cm has 660 turns and a length of 25.0 cm. (a) Find its

Question

A solenoid of radius 1.50 cm has 660 turns and a length of 25.0 cm. (a) Find its inductance. mH (b) Find the rate at which current must change through it to produce an emf of 80.0 mV. (Enter the magnitude.) A/s A 35.5 mA current is carried by a uniformly wound air-core solenoid with 455 turns, a 18.5 mm diameter, and 11.5 cm length (a) Compute the magnetic field inside the solenoid. mu T (b) Compute the magnetic flux through each turn. T middot m^2 (c) Compute the inductance of the solenoid. mH (d) Which of these quantities depends on the current? (Select all that apply.) magnetic field inside the solenoid magnetic flux through each turn inductance of the solenoid

Explanation / Answer

1)

a)

L = µ*N^2*A / d

L = inductance Henry

µ = the absolute magnetic permeability (µo = 4pi x 10^-7 H/m).

N = 455 turns

length in meter = 0.25 m

A = pi(r)^2 = pi(1.5 x 10^-2)^2 = 7.065 x 10^-4 m^2

L = (4pi x 10^-7 x 455^2 x 7.065 x 10^-4) / 0.25

L = 0.735 mH

b)

emf = - L(delta I /delta t)

80 x 10^-3 = 0.735 x 10^-3 ( delta I /delta t)

delta I / delta t = 108.84 A/s

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