Question

A solenoid 4.0 cm in radius and 4.0 m in length has 8000 uniformly spaced turns and carries a current of 5.0 A. Consider a plane circular surface (radius = 2.0 cm) located at the center of the solenoid with its axis coincident with the axis of the solenoid. What is the magnetic flux through this surface? (1 Wb = 1 T ⋅ m2)Group of answer choices

Answer 1

Answer:

Magnetic flux, $\phi=1.57\times 10^{-5}\ Wb$

Explanation:

It is given that,

Radius of the solenoid, r = 4 cm = 0.04 m

Length of the solenoid, L = 4 m

No of turns, N = 8000

Current, I = 5 A

Radius of the plane circular surface, r' = 2 cm = 0.02

Area of the circular surface,

$A=\pi r'^2$

$A=\pi (0.02)^2=0.00125\ m^2$                  

The magnetic flux through this surface is given by :

$\phi=B\times A$

B is the magnetic field of the solenoid

$\phi=\mu_o\dfrac{N}{L}I\times A$

$\phi=4\pi \times 10^{-7}\times \dfrac{8000}{4}\times 5\times 0.00125$

$\phi=1.57\times 10^{-5}\ Wb$

So, the magnetic flux through this surface is $1.57\times 10^{-5}\ Wb$/. Hence, this is the required solution.