A 310 turn solenoid with a length of 18.0 cm and a radius of 1.60 cm carries a current of 1.90 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 310 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 310 turn solenoid. T(b) Calculate the magnetic field of the 310 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. VIs it equal to the instantaneous induced emf? Explain.(f) Why could contributions to the magnetic field by the current in the 4-turn coil be neglected in this calculation?

Given that;

Number of turns in the solenoid N = 310

Length of the solenoid L = 18 cm = 0.18 m

Radius of the solenoid r = 1.60 cm = 0.016 m

Current in the first Circuit I₁ = 1.90A

Number of turns in second coil N₂ = 4

Final Current solenoid I₂ = 5.0 A

Time interval to change the time Δt = 0.9 s

a)

According to Ampere's law, magnetic field inside a conductor is calculated as;

B₁ = ц₀N₁I₁ / L

(ц₀ = 4π × 10⁻⁷  constant)

therefore we substitute

{(4π × 10⁻⁷) ×  310 × 1.9A} / 0.18m

= 0.0041 T

b)

Magnetic field inside the solenoid after t = 0.9

B₁ = ц₀N₁I₂ / L

= {(4π × 10⁻⁷) ×  310 × 5.0A} / 0.18m

= 0.0108 T

c)

Area of coil is

A = πr²

A = π ×  ( 0.016 )²

A = 0.000804 m²

d)

Change in magnetic influx is

dФ = (B₂ - B₁) A

= ( 0.0108 T - 0.0041 T) × 0.000804 m²

= 0.0000053868 ≈ 5.39 × 10⁻⁶

e)

Average induced emf is

e = -N₂ ( dФ / dt )

e = ( -4 ) ( 5.39 × 10⁻⁶ / 0.9)

e = - 2.39 × 10⁻⁵V ( NOTE, this is not equal to the instantaneous induced emf )

f)

The induced emf is very low, so the contributions to the magnetic field in the coil is Negative.