A solenoid 4cm in diameter and 20 cm in length has 250 turns and carries a current of 15A. Calculate the flux through the surface of a disc of 10 cm radius that is positioned perpendicular to and centered on the axis of the solenoid.
A solenoid 4cm in diameter and 20 cm in length has 250 turns and carries a current of 15A. Calculate the flux through the surface of a disc of 10 cm radius that is positioned perpendicular to and centered on the axis of the solenoid.
Step 1, Given data
R = Radius of solenoid which is = 2 cm
Length, L= 20 cm
No of turns= N=250
Turns per unit length=n=1250 turns/meter
current I = 15 A
Step 2, Finding the flux
We know,
Magnetic Flux = BAcosθ
Here the angle between the magnetic field and the area is θ=00
We know cos00=1
Therefore
Flux = B A cos00 = B A
Now, we know
B = μ₀ n i
Putting all the values
B = 4π×10−7×1250×15
The area is given as
A = πr² = π×2×2×10−4 (converting centimeter to meter)
After solving
B = 2.356×10−2 T
A =12.56×10−4 m2
Now we can write
Flux =2.356×10−2×12.56×10−4 = 29.6×10−6 Wb
Hence the flux is 29.6×10−6Wb
Step 1, Given data
R = Radius of solenoid which is = 2 cm
Length, L= 20 cm
No of turns= N=250
Turns per unit length=n=1250 turns/meter
current I = 15 A
Step 2, Finding the flux
We know,
Magnetic Flux = BAcosθ
Here the angle between the magnetic field and the area is θ=00
We know cos00=1
Therefore
Flux = B A cos00 = B A
Now, we know
B = μ₀ n i
Putting all the values
B = 4π×10−7×1250×15
The area is given as
A = πr² = π×2×2×10−4 (converting centimeter to meter)
After solving
B = 2.356×10−2 T
A =12.56×10−4 m2
Now we can write
Flux =2.356×10−2×12.56×10−4 = 29.6×10−6 Wb
Hence the flux is 29.6×10−6Wb
