A solenoid has 2000turns wound over a length of .3m Its cross sectional area is .0012(m)(m).around its central section a coil of 300 turns is wound .If an initial current of 2 amp flow in solenoid is reversed in .25sec. The emf induced in CPI will be what?
To calculate the emf induced in the coil, we can use Faraday's law of electromagnetic induction, which states that the emf (ε) induced in a coil is equal to the rate of change of magnetic flux through the coil.
Step 1: Calculate the rate of change of magnetic flux (dΦ/dt):
We know that the magnetic flux (Φ) is given by the product of the magnetic field (B) and the cross-sectional area (A) of the coil:
Φ = B * A
Since the solenoid has 2000 turns and a length of 0.3m, the total number of turns in the solenoid is:
N = 2000
The total length of the solenoid is 0.3m, which means the length per turn is:
l = 0.3m / 2000 = 0.00015m
Step 2: Calculate the magnetic field (B):
The magnetic field inside a solenoid can be approximated by the equation:
B = μ * N * I / l
Where:
μ is the permeability of free space, approximately equal to 4π * 10^(-7) T*m/A
N is the number of turns
I is the current flowing through the solenoid
l is the length per turn
Given:
N = 2000
I = 2A
l = 0.00015m
We can substitute these values into the equation to find the magnetic field:
B = (4π * 10^(-7) T*m/A) * 2000 * 2A / 0.00015m = 0.033508 T
Step 3: Calculate the cross-sectional area (A) of the coil:
The cross-sectional area of the coil is given as 0.0012 (m)(m).
Step 4: Calculate the rate of change of magnetic flux (dΦ/dt):
dΦ/dt = B * A / dt
Given that the current is reversed in 0.25 sec, we have:
dt = 0.25 sec
Substituting the values into the equation, we get:
dΦ/dt = 0.033508 T * 0.0012 m^2 / 0.25 s = 0.160824 V
Therefore, the emf induced in the coil will be approximately 0.160824 volts.
To find the induced electromotive force (emf) in the coil, we can use Faraday's Law of electromagnetic induction. The equation for the emf induced is given by:
emf = -N(dΦ/dt)
Where:
- emf is the induced electromotive force,
- N is the number of turns in the coil (300 turns),
- dΦ/dt is the rate of change of magnetic flux.
To calculate dΦ/dt, we need to find the change in magnetic flux (dΦ) over the change in time (dt).
The magnetic flux (Φ) through the coil is given by:
Φ = B * A
Where:
- B is the magnetic field strength,
- A is the cross-sectional area of the coil (0.0012 m^2).
The magnetic field strength (B) inside the solenoid is given by:
B = μ₀ * (N₁ * I₁ + N₂ * I₂) / L
Where:
- μ₀ is the permeability of free space (constant),
- N₁ is the number of turns in the solenoid (2000 turns),
- I₁ is the initial current (2 A),
- N₂ is the number of turns in the coil (300 turns),
- I₂ is the final current (reversed direction),
- L is the length of the solenoid (0.3 m).
Substituting the values into the equation, we get:
B = μ₀ * (2000 * 2 + 300 * (-2)) / 0.3
Next, calculate the change in magnetic flux (dΦ) by subtracting the initial flux from the final flux:
dΦ = Φ₂ - Φ₁
Substituting the values into the equation, we get:
dΦ = B₂ * A - B₁ * A
Finally, calculate the rate of change of magnetic flux (dΦ/dt) by dividing the change in magnetic flux by the time interval:
dΦ/dt = dΦ / dt
Substituting the values into the equation, we get:
dΦ/dt = (B₂ * A - B₁ * A) / dt
Now, substitute the calculated value of dΦ/dt into the formula for emf:
emf = -N * dΦ/dt
Substituting the values, calculate the value of emf.