Question

A solenoid of radius r 5 1.25 cm and length , 5 30.0 cm has 300 turns and carries 12.0 A. (a) Calculate the flux through the surface of a disk-shaped area of radius R 5 5.00 cm that is positioned perpendicular to and centered on the axis of the solenoid as

Answer 1

Answer:

$\phi = 7.4 \times 10^{-6} Wb$

Explanation:

Magnetic field due to a long solenoid at the center is given by

$B = \frac{\mu_o N i}{L}$

here we know that

N = 300

L = 30 cm

i = 12 A

now magnetic field is given as

$B = \frac{(4\pi \times 10^{-7})(300)(12)}{0.30}$

$B = 0.015 T$

Now magnetic flux through the disc is given as

$\phi = B.A$

$\phi = (0.015)(\pi r^2)$

$\phi = (0.015)(\pi)(0.0125)^2$

$\phi = 7.4 \times 10^{-6} Wb$