Question

A long solenoid has turns per unit length 1.5 x 10³ per meter and area of cross-section 25 cm². A coil C of 150 turns is wound tightly around the center of the solenoid. For a current of 3 A in the solenoid, calculate
(a) the magnetic flux density at the centre of the solenoid
(b) the flux linkage in the coil C
(c) the average emf induced in coil C if the current in the solenoid is reversed in the direction in a time of 0.5 s. [µ₀ = 4π x 10^-7 H/m]

Answer 1

Data: n = 1.5 x 10³ m^-1, A = 25 x 10^-4 m², 

N_C = 150, I = 3A, Δ t = 0.5 s,

µ₀ = 4π x 10^-7 H/m

(a) Magnetic flux density inside the solenoid,

B = `mu_0 "nI" = (4pi xx 10^-7)(1500)(3)`

= 5.656 × 10^-3 T = 5.656 mT

(b) Flux per unit turn through the coils of the solenoid, Φ_m = BA

Since the coil C is wound tightly over the solenoid, the flux linkage of C is

N_CΦ_m = N_CBA = (150)(5.656 × 10^-3)(25 × 10^-4)

= 2.121 × 10^-3 Wb = 2.121 mWb

(c) Initial flux through coil C,

Φ_i = N_CΦ_m = 2.121 × 10^-3 Wb

Reversing the current in the solenoid reverses the flux through coil C, the magnitude remaining the same. But since the flux enters through the other face of the coil, the final flux through C is 

Φ_i = - 2.121 × 10^-3 Wb

Therefore, the average emf induced in coil C,

e = `(phi_"f" - phi_"i")/(triangle"t")`

`= ((- 2.121 - 2.121) xx 10^-3)/0.5`

= 2 × 4.242 × 10^-3 = 8.484 × 10^-3 V = 8.484 mV