A very long, straight solenoid with a cross-sectional area of 1.90 cm2 is wound with 86.6 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i (t) = (0.176 A/s2) t2. A secondary winding of 5.0 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A?

Question

Physics

Bullock

26 July, 07:31

A very long, straight solenoid with a cross-sectional area of 1.90 cm2 is wound with 86.6 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i (t) = (0.176 A/s2) t2. A secondary winding of 5.0 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A?

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Answers (1)

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Roy Velasquez · Answer 1

|Emf| = 1.58x10⁻⁵ V

Explanation:

N = 88.6 turns/cm

N = 8860 turns/metre

The field inside the long solenoid is given by

B = μ₀Ni

B = 4πx10⁻⁷ x 8860 x 0.1761t² =

B = 1.96x10⁻³ t²

dB/dt = 3.92x10⁻³ t

A = 1.90cm² = 1.90x10⁻⁴ m²

|Emf| = rate of change of flux linkage

|Emf| = d (NAB) / dt

|Emf| = NA dB/dt

|Emf| = 5 x 1.90x10⁻⁴ * 3.92x10⁻³ t

|Emf| = 3.724x10⁻⁶ t

if T is the time at which the current = 3.2A

3.2 A = 0.176T²

T = 4.26s

|Emf| = 3.724x10⁻⁶t * 4.26s

|Emf| = 1.58x10⁻⁵ V