Question

When a current of 0.05 A is passed though a solenoid of length 10 cm, an axial magnetic field of 40.5 Gauss is geneated inside it. Now, the same solenoid is used as the primary coil in a transformer. The power supply linked to the transformer is at 220 V, 50 Hz. What should be the number of turns in the secondary coil to operate a device at 10 V, 50 Hz? Show all calculations. Not showing calculations will get points deducted.

Answer 1

212.2

Step-by-step explanation:

because it is

Answer 2

Ns = 293 turns

Step-by-step explanation:

The magnetic field of a solenoid is given as follows:

B = μ₀NI/L

where,

B = Magnetic Field = (40.5 Gauss)(1 T/10000 Gauss) = 4.05 x 10⁻³ T

μ₀ = Permeability of Free Space = 4π x 10⁻⁷ T/Am

N = No. turns in solenoid = ?

I = Current = 0.05 A

L = Length = 10 cm = 0.1 m

Therefore,

4.05 x 10⁻³ = (4π x 10⁻⁷ )(N)(0.05)/0.1

N = (4.05 x 10⁻³)/(6.28 x 10⁻⁷)

N = 6446 turns

Now, we use turns ratio formula for transformer:

Np/Ns = Vp/Vs

where,

Np = No. of turns in primary coil = 6446 turns

Ns = No. of turns in secondary coil = ?

Vp = Primary Voltage = 220 V

Vs = Secondary Voltage = 10 V

Therefore,

6446 turns/Ns = 220 V/10 V

Ns = 64460 turns/220 V

Ns = 293 turns