Along solenoid is wound with 500 turns permeter and the current in its windings is increasing at rate of 20A/s .If the cross sectional area of solenoid 5cm^2.what will be the rate change of magnetic flux
To find the rate of change of magnetic flux through the solenoid, we can use Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) in a circuit is equal to the rate of change of magnetic flux through the circuit.
The magnetic flux through the solenoid can be calculated by multiplying the magnetic field strength (B) by the cross-sectional area (A) of the solenoid and the number of turns (N) per unit length.
Magnetic Flux (Φ) = B * A * N
Now, let's find the rate of change of magnetic flux (dΦ/dt) by differentiating the above equation with respect to time (t).
dΦ/dt = (dB/dt) * A * N
Given that the current in the solenoid is increasing at a rate of 20 A/s, we can relate it to the magnetic field strength using Ampere's law for a solenoid.
B = μ₀ * (N/L) * I
Where:
μ₀ = permeability of free space (constant)
N = number of turns
L = length of the solenoid
I = current through the solenoid
Since we're given the number of turns per meter, and the cross-sectional area is given, we can assume the length of the solenoid is 1 meter.
Now, let's substitute the value of B in the previous equation.
dΦ/dt = (d/dt) (μ₀ * (N/L) * I) * A * N
Since the length is given as 1 meter, the equation simplifies to:
dΦ/dt = (d/dt) (μ₀ * N² * I) * A
Finally, substitute the given values:
N = 500 turns per meter
I = 20 A/s
A = 5 cm² = 5 * 10⁻⁴ m²
μ₀ = 4π × 10⁻⁷ T·m/A
Now, plug in these values and calculate dΦ/dt to find the rate of change of magnetic flux through the solenoid.