Question

A 280 turn solenoid with a length of 21.0 cm and a radius of 1.80 cm carries a current of 2.10 A. A second coil of four turns is wrapped tightly around this solenoid, so it can be considered to have the same radius as the solenoid. The current in the 280 turn solenoid increases steadily to 5.00 A in 0.900 s.

(a) Use Ampere's law to calculate the initial magnetic field in the middle of the 280 turn solenoid.
T

(b) Calculate the magnetic field of the 280 turn solenoid after 0.900 s.
T

(c) Calculate the area of the 4-turn coil.
m²

(d) Calculate the change in the magnetic flux through the 4-turn coil during the same period.
Wb

(e) Calculate the average induced emf in the 4-turn coil.
V

Answer 1

No of turns in solenoid , N = 280 turns

Length of solenoid , L = 21 cm = 0.21 m

Radius of solenoid, r = 1.80 cm = 0.018 m

Initial current in solenoid , I₁ = 2.10 A

final current in solenoid, I₂ = 5A

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a) Magnetic field inside the solenoid , B₁ = u₀ n I₁   ( n = no of turns per unit length, N = n*L)

B1 = u₀ ( N/ L) * I₁ ( n = N / L)

= ( 4*pie* 10^-7 Tm/A) * ( 280) * (2.10) / 0.21

= 3.516 * 10^-3 T

b) mag. field of 280 turns solenoid after 0.900 s

B₂ = u₀* n* I₂

= u₀ * ( N/ L) * I₂

= (4*pie*10^-7) * ( 280) * 5 / 0.21

= 8.37*10^-3 T

c) area of 4 turn coli

A = pie* r²

= 3.14* ( 0.018)²

= 10.17*10^-4 m²

d) change in mag. flux(d@) through 4 turn coli

d@ = ( B₂ - B₁ ) *A

= ( 8.37*10^-3 - 3.516*10^-3 ) * 10.17*10^-4

= 4.936*10^-6 Wb

e) average induced emf in 4-turn coil

e = - N' d@ / dt

= - 4* 4.936*10^-6 / 0.900

e = - 2.193*10^-5 V

here -ve sign is for direction of induced emf.